Dalam permainan Flood Paint, tujuan dari permainan ini adalah untuk membuat seluruh papan menjadi warna yang sama dalam beberapa putaran mungkin.

Gim dimulai dengan papan yang terlihat seperti ini:

3 3 5 4 1 3 4 1 5
5 1 3 4 1 1 5 2 1
6 5 2 3 4 3 3 4 3
4 4 4 5 5 5 4 1 4
6 2 5 3[3]1 1 6 6
5 5 1 2 5 2 6 6 3
6 1 1 5 3 6 2 3 6
1 2 2 4 5 3 5 1 2
3 6 6 1 5 1 3 2 4

Saat ini, angka (mewakili warna) di tengah papan adalah 3. Setiap belokan, persegi di pusat akan berubah warna, dan semua kotak dengan warna yang sama yang dapat dijangkau dari pusat dengan bergerak secara horizontal atau vertikal ( yaitu di wilayah banjir pusat alun-alun) akan berubah warna dengan itu. Jadi jika alun-alun pusat berubah warna menjadi 5:

3 3 5 4 1 3 4 1 5
5 1 3 4 1 1 5 2 1
6 5 2 3 4 3 3 4 3
4 4 4 5 5 5 4 1 4
6 2 5 5[5]1 1 6 6
5 5 1 2 5 2 6 6 3
6 1 1 5 3 6 2 3 6
1 2 2 4 5 3 5 1 2
3 6 6 1 5 1 3 2 4

maka 3 yang ada di sebelah kiri tengah 3 juga akan berubah warna. Sekarang ada total tujuh 5 yang dapat dicapai dari satu pusat, dan jadi jika kita kemudian berubah warna menjadi 4:

3 3 5 4 1 3 4 1 5
5 1 3 4 1 1 5 2 1
6 5 2 3 4 3 3 4 3
4 4 4 4 4 4 4 1 4
6 2 4 4[4]1 1 6 6
5 5 1 2 4 2 6 6 3
6 1 1 5 3 6 2 3 6
1 2 2 4 5 3 5 1 2
3 6 6 1 5 1 3 2 4

daerah yang dicat kembali bertambah besar secara dramatis.

Tugas Anda adalah membuat program yang akan mengambil kisi warna 19-by-19 dari 1 hingga 6 sebagai input, dalam bentuk apa pun yang Anda pilih:

4 5 1 1 2 2 1 6 2 6 3 4 2 3 2 3 1 6 3
4 2 6 3 4 4 5 6 4 4 5 3 3 3 3 5 4 3 4
2 3 5 2 2 5 5 1 2 6 2 6 6 2 1 6 6 1 2
4 6 5 5 5 5 4 1 6 6 3 2 6 4 2 6 3 6 6
1 6 4 4 4 4 6 4 2 5 5 3 2 2 4 1 5 2 5
1 6 2 1 5 1 6 4 4 1 5 1 3 4 5 2 3 4 1
3 3 5 3 2 2 2 4 2 1 6 6 6 6 1 4 5 2 5
1 6 1 3 2 4 1 3 3 4 6 5 1 5 5 3 4 3 3
4 4 1 5 5 1 4 6 3 3 4 5 5 6 1 6 2 6 4
1 4 2 5 6 5 5 3 2 5 5 5 3 6 1 4 4 6 6
4 6 6 2 6 6 2 4 2 6 1 5 6 2 3 3 4 3 6
6 1 3 6 3 5 5 3 6 1 3 4 4 5 1 2 6 4 3
2 6 1 3 2 4 2 6 1 1 5 2 6 6 6 6 3 3 3
3 4 5 4 6 6 3 3 4 1 1 6 4 5 1 3 4 1 2
4 2 6 4 1 5 3 6 4 3 4 5 4 2 1 1 4 1 1
4 2 4 1 5 2 2 3 6 6 6 5 2 5 4 5 4 5 1
5 6 2 3 4 6 5 4 1 3 2 3 2 1 3 6 2 2 4
6 5 4 1 3 2 2 1 1 1 6 1 2 6 2 5 6 4 5
5 1 1 4 2 6 2 5 6 1 3 3 4 1 6 1 2 1 2

dan kembalikan urutan warna yang akan diubah pusat persegi untuk setiap belokan, sekali lagi dalam format yang Anda pilih:

263142421236425431645152623645465646213545631465

Pada akhir setiap urutan gerakan, kotak dalam kisi 19-by-19 semuanya harus memiliki warna yang sama.

Program Anda harus sepenuhnya bersifat deterministik; solusi pseudorandom diperbolehkan, tetapi program harus menghasilkan output yang sama untuk test case yang sama setiap waktu.

Program yang menang akan mengambil jumlah langkah paling sedikit untuk menyelesaikan semua 100.000 kasus uji yang ditemukan dalam file ini (file teks zip, 14,23 MB). Jika dua solusi mengambil jumlah langkah yang sama (misalnya jika mereka berdua menemukan strategi yang optimal), program yang lebih pendek akan menang.

BurntPizza telah menulis sebuah program di Jawa untuk memverifikasi hasil pengujian. Untuk menggunakan program ini, jalankan kiriman Anda dan kirimkan output ke file bernama steps.txt. Kemudian, jalankan program ini dengan steps.txtdan floodtestfile di direktori yang sama. Jika entri Anda valid dan menghasilkan solusi yang benar untuk semua file, itu harus lulus semua tes dan kembaliAll boards solved successfully.

import java.io.*;
import java.util.*;

public class PainterVerifier {

    public static void main(String[] args) throws FileNotFoundException {

        char[] board = new char[361];

        Scanner s = new Scanner(new File("steps.txt"));
        Scanner b = new Scanner(new File("floodtest"));

        int lineNum = 0;

        caseloop: while (b.hasNextLine()) {

            for (int l = 0; l < 19; l++) {
                String lineb = b.nextLine();
                if (lineb.isEmpty())
                    continue caseloop;
                System.arraycopy(lineb.toCharArray(), 0, board, l * 19, 19);
            }

            String line = s.nextLine();
            if (line.isEmpty())
                continue;
            char[] steps = line.toCharArray();

            Stack<Integer> nodes = new Stack<Integer>();

            for (char c : steps) {
                char targetColor = board[180];
                char replacementColor = c;

                nodes.push(180);

                while (!nodes.empty()) {
                    int n = nodes.pop();
                    if (n < 0 || n > 360)
                        continue;
                    if (board[n] == targetColor) {
                        board[n] = replacementColor;
                        if (n % 19 > 0)
                            nodes.push(n - 1);
                        if (n % 19 < 18)
                            nodes.push(n + 1);
                        if (n / 19 > 0)
                            nodes.push(n - 19);
                        if (n / 19 < 18)
                            nodes.push(n + 19);
                    }
                }
            }
            char center = board[180];
            for (char c : board)
                if (c != center) {
                    s.close();
                    b.close();

                    System.out.println("\nIncomplete board found!\n\tOn line " + lineNum + " of steps.txt");
                    System.exit(0);
                }

            if (lineNum % 5000 == 0)
                System.out.printf("Verification %d%c complete...\n", lineNum * 100 / 100000, '%');

            lineNum++;
        }
        s.close();
        b.close();
        System.out.println("All boards solved successfully.");
    }
}

Juga, papan skor, karena hasilnya tidak benar-benar diurutkan berdasarkan skor dan di sini itu sangat penting:

1. 1.985.078 - smack42, Jawa
2. 2.075.452 - user1502040, C
3. 2.098.382 - tigrou, C #
4. 2.155.834 - CoderTao, C #
5. 2.201.995 - MrBackend, Jawa
6. 2.383.569 - CoderTao, C #
7. 2.384.020 - Herjan, C
8. 2.403.189 - Origineil, Jawa
9. 2.445.761 - Herjan, C
10. 2.475.056 - Daftar Jeremy, Haskell
11. 2.480.714 - SteelTermite, C (2.395 byte)
12. 2.480.714 - Herjan, Jawa (4,702 bytes)
13. 2.588.847 - BurntPizza, Java (2,748 bytes)
14. 2.588.847 - Gero3, node.js (4,641 bytes)
15. 2.979.145 - Teun Pronk, Delphi XE3
16. 4.780.841 - BurntPizza, Jawa
17. 10.800.000 - Joe Z., Python

Java - 1.985.078 langkah

https://github.com/smack42/ColorFill

Entri terlambat lainnya. File hasil yang berisi langkah-langkah 1.985.078 dapat ditemukan di sini .

Beberapa info latar belakang:

Saya menemukan tantangan ini beberapa tahun yang lalu, ketika saya mulai memprogram klon saya sendiri dari game Flood-it.

Algoritma DFS dan A * "terbaik-tidak lengkap"
Sejak awal, saya ingin membuat algoritma pemecah yang baik untuk game ini. Seiring waktu, saya dapat memperbaiki solver saya dengan memasukkan beberapa strategi yang melakukan pencarian tidak lengkap berbeda (mirip dengan yang digunakan dalam program lain di sini) dan dengan menggunakan hasil terbaik dari strategi tersebut untuk setiap solusi. Saya bahkan menerapkan kembali algoritma A * tigrou di Java dan menambahkannya ke solver saya untuk mencapai solusi keseluruhan yang lebih baik daripada hasil tigrou.

Algoritma DFS lengkap
Lalu saya fokus pada algoritma yang selalu menemukan solusi optimal. Saya menghabiskan banyak upaya untuk mengoptimalkan strategi pencarian mendalam-pertama saya. Untuk mempercepat pencarian, saya menyertakan peta hash yang menyimpan semua status yang dieksplorasi, sehingga pencarian dapat menghindari menjelajahinya lagi. Meskipun algoritma ini berfungsi dengan baik dan menyelesaikan semua 14x14 teka-teki dengan cukup cepat, ia menggunakan terlalu banyak memori dan berjalan sangat lambat dengan 19x19 teka-teki dalam tantangan kode ini.

Algoritma Puchert A *
Beberapa bulan yang lalu saya dihubungi untuk melihat pemecah Flood-It oleh Aaron dan Simon Puchert . Program itu menggunakan algoritma tipe A * dengan heuristik yang dapat diterima (berbeda dengan tigrou) dan memindahkan pemangkasan yang mirip dengan Pencarian Titik-Langsung. Saya dengan cepat memperhatikan bahwa program ini sangat cepat dan menemukan solusi optimal !

Tentu saja, saya harus mengimplementasikan kembali algoritma yang hebat ini dan menambahkannya ke program saya. Saya berupaya mengoptimalkan program Java saya agar berjalan secepat program C ++ asli oleh saudara-saudara Puchert. Kemudian saya memutuskan untuk mencoba 100.000 uji kasus dari tantangan ini. Di mesin saya, program berjalan selama lebih dari 120 jam untuk menemukan 1.985.078 langkah, menggunakan implementasi saya dari algoritma Puchert A * .

Ini menjadi solusi terbaik untuk tantangan ini, kecuali ada beberapa bug dalam program yang akan menghasilkan solusi yang kurang optimal. Umpan balik apapun diterima!

sunting: menambahkan bagian kode yang relevan di sini:

kelas AStarPuchertStrategy

/**
 * a specific strategy for the AStar (A*) solver.
 * <p>
 * the idea is taken from the program "floodit" by Aaron and Simon Puchert,
 * which can be found at <a>https://github.com/aaronpuchert/floodit</a>
 */
public class AStarPuchertStrategy implements AStarStrategy {

    private final Board board;
    private final ColorAreaSet visited;
    private ColorAreaSet current, next;
    private final short[] numCaNotFilledInitial;
    private final short[] numCaNotFilled;

    public AStarPuchertStrategy(final Board board) {
        this.board = board;
        this.visited = new ColorAreaSet(board);
        this.current = new ColorAreaSet(board);
        this.next = new ColorAreaSet(board);
        this.numCaNotFilledInitial = new short[board.getNumColors()];
        for (final ColorArea ca : board.getColorAreasArray()) {
            ++this.numCaNotFilledInitial[ca.getColor()];
        }
        this.numCaNotFilled = new short[board.getNumColors()];
    }

    /* (non-Javadoc)
     * @see colorfill.solver.AStarStrategy#setEstimatedCost(colorfill.solver.AStarNode)
     */
    @Override
    public void setEstimatedCost(final AStarNode node) {

        // quote from floodit.cpp: int State::computeValuation()
        // (in branch "performance")
        //
        // We compute an admissible heuristic recursively: If there are no nodes
        // left, return 0. Furthermore, if a color can be eliminated in one move
        // from the current position, that move is an optimal move and we can
        // simply use it. Otherwise, all moves fill a subset of the neighbors of
        // the filled nodes. Thus, filling that layer gets us at least one step
        // closer to the end.

        node.copyFloodedTo(this.visited);
        System.arraycopy(this.numCaNotFilledInitial, 0, this.numCaNotFilled, 0, this.numCaNotFilledInitial.length);
        {
            final ColorAreaSet.FastIteratorColorAreaId iter = this.visited.fastIteratorColorAreaId();
            int nextId;
            while ((nextId = iter.nextOrNegative()) >= 0) {
                --this.numCaNotFilled[this.board.getColor4Id(nextId)];
            }
        }

        // visit the first layer of neighbors, which is never empty, i.e. the puzzle is not solved yet
        node.copyNeighborsTo(this.current);
        this.visited.addAll(this.current);
        int completedColors = 0;
        {
            final ColorAreaSet.FastIteratorColorAreaId iter = this.current.fastIteratorColorAreaId();
            int nextId;
            while ((nextId = iter.nextOrNegative()) >= 0) {
                final byte nextColor = this.board.getColor4Id(nextId);
                if (--this.numCaNotFilled[nextColor] == 0) {
                    completedColors |= 1 << nextColor;
                }
            }
        }
        int distance = 1;

        while(!this.current.isEmpty()) {
            this.next.clear();
            final ColorAreaSet.FastIteratorColorAreaId iter = this.current.fastIteratorColorAreaId();
            int thisCaId;
            if (0 != completedColors) {
                // We can eliminate colors. Do just that.
                // We also combine all these elimination moves.
                distance += Integer.bitCount(completedColors);
                final int prevCompletedColors = completedColors;
                completedColors = 0;
                while ((thisCaId = iter.nextOrNegative()) >= 0) {
                    final ColorArea thisCa = this.board.getColorArea4Id(thisCaId);
                    if ((prevCompletedColors & (1 << thisCa.getColor())) != 0) {
                        // completed color
                        // expandNode()
                        for (final int nextCaId : thisCa.getNeighborsIdArray()) {
                            if (!this.visited.contains(nextCaId)) {
                                this.visited.add(nextCaId);
                                this.next.add(nextCaId);
                                final byte nextColor = this.board.getColor4Id(nextCaId);
                                if (--this.numCaNotFilled[nextColor] == 0) {
                                    completedColors |= 1 << nextColor;
                                }
                            }
                        }
                    } else {
                        // non-completed color
                        // move node to next layer
                        this.next.add(thisCaId);
                    }
                }
            } else {
                // Nothing found, do the color-blind pseudo-move
                // Expand current layer of nodes.
                ++distance;
                while ((thisCaId = iter.nextOrNegative()) >= 0) {
                    final ColorArea thisCa = this.board.getColorArea4Id(thisCaId);
                    // expandNode()
                    for (final int nextCaId : thisCa.getNeighborsIdArray()) {
                        if (!this.visited.contains(nextCaId)) {
                            this.visited.add(nextCaId);
                            this.next.add(nextCaId);
                            final byte nextColor = this.board.getColor4Id(nextCaId);
                            if (--this.numCaNotFilled[nextColor] == 0) {
                                completedColors |= 1 << nextColor;
                            }
                        }
                    }
                }
            }

            // Move the next layer into the current.
            final ColorAreaSet tmp = this.current;
            this.current = this.next;
            this.next = tmp;
        }
        node.setEstimatedCost(node.getSolutionSize() + distance);
    }

}

bagian dari kelas AStarSolver

private void executeInternalPuchert(final ColorArea startCa) throws InterruptedException {
    final Queue<AStarNode> open = new PriorityQueue<AStarNode>(AStarNode.strongerComparator());
    open.offer(new AStarNode(this.board, startCa));
    AStarNode recycleNode = null;
    while (open.size() > 0) {
        if (Thread.interrupted()) { throw new InterruptedException(); }
        final AStarNode currentNode = open.poll();
        if (currentNode.isSolved()) {
            this.addSolution(currentNode.getSolution());
            return;
        } else {
            // play all possible colors
            int nextColors = currentNode.getNeighborColors(this.board);
            while (0 != nextColors) {
                final int l1b = nextColors & -nextColors; // Integer.lowestOneBit()
                final int clz = Integer.numberOfLeadingZeros(l1b); // hopefully an intrinsic function using instruction BSR / LZCNT / CLZ
                nextColors ^= l1b; // clear lowest one bit
                final byte color = (byte)(31 - clz);
                final AStarNode nextNode = currentNode.copyAndPlay(color, recycleNode, this.board);
                if (null != nextNode) {
                    recycleNode = null;
                    this.strategy.setEstimatedCost(nextNode);
                    open.offer(nextNode);
                }
            }
        }
        recycleNode = currentNode;
    }
}

bagian dari kelas AStarNode

/**
 * check if this color can be played. (avoid duplicate moves)
 * the idea is taken from the program "floodit" by Aaron and Simon Puchert,
 * which can be found at <a>https://github.com/aaronpuchert/floodit</a>
 * @param nextColor
 * @return
 */
private boolean canPlay(final byte nextColor, final List<ColorArea> nextColorNeighbors) {
    final byte currColor = this.solution[this.solutionSize];
    // did the previous move add any new "nextColor" neighbors?
    boolean newNext = false;
next:   for (final ColorArea nextColorNeighbor : nextColorNeighbors) {
        for (final ColorArea prevNeighbor : nextColorNeighbor.getNeighborsArray()) {
            if ((prevNeighbor.getColor() != currColor) && this.flooded.contains(prevNeighbor)) {
                continue next;
            }
        }
        newNext = true;
        break next;
    }
    if (!newNext) {
        if (nextColor < currColor) {
            return false;
        } else {
            // should nextColor have been played before currColor?
            for (final ColorArea nextColorNeighbor : nextColorNeighbors) {
                for (final ColorArea prevNeighbor : nextColorNeighbor.getNeighborsArray()) {
                    if ((prevNeighbor.getColor() == currColor) && !this.flooded.contains(prevNeighbor)) {
                        return false;
                    }
                }
            }
            return true;
        }
    } else {
        return true;
    }
}

/**
 * try to re-use the given node or create a new one
 * and then play the given color in the result node.
 * @param nextColor
 * @param recycleNode
 * @return
 */
public AStarNode copyAndPlay(final byte nextColor, final AStarNode recycleNode, final Board board) {
    final List<ColorArea> nextColorNeighbors = new ArrayList<ColorArea>(128);  // constant, arbitrary initial capacity
    final ColorAreaSet.FastIteratorColorAreaId iter = this.neighbors.fastIteratorColorAreaId();
    int nextId;
    while ((nextId = iter.nextOrNegative()) >= 0) {
        final ColorArea nextColorNeighbor = board.getColorArea4Id(nextId);
        if (nextColorNeighbor.getColor() == nextColor) {
            nextColorNeighbors.add(nextColorNeighbor);
        }
    }
    if (!this.canPlay(nextColor, nextColorNeighbors)) {
        return null;
    } else {
        final AStarNode result;
        if (null == recycleNode) {
            result = new AStarNode(this);
        } else {
            // copy - compare copy constructor
            result = recycleNode;
            result.flooded.copyFrom(this.flooded);
            result.neighbors.copyFrom(this.neighbors);
            System.arraycopy(this.solution, 0, result.solution, 0, this.solutionSize + 1);
            result.solutionSize = this.solutionSize;
            //result.estimatedCost = this.estimatedCost;  // not necessary to copy
        }
        // play - compare method play()
        for (final ColorArea nextColorNeighbor : nextColorNeighbors) {
            result.flooded.add(nextColorNeighbor);
            result.neighbors.addAll(nextColorNeighbor.getNeighborsIdArray());
        }
        result.neighbors.removeAll(result.flooded);
        result.solution[++result.solutionSize] = nextColor;
        return result;
    }
}

C # - 2.098.382 langkah

Saya mencoba banyak hal, kebanyakan dari mereka gagal dan tidak berfungsi sama sekali, sampai saat ini. Saya mendapat sesuatu yang cukup menarik untuk mengirim jawaban.

Pasti ada cara untuk meningkatkan ini lebih jauh. Saya pikir akan di bawah langkah 2M mungkin.

Butuh kira-kira 7 hoursuntuk menghasilkan hasil. Ini adalah file txt dengan semua solusi, jika seseorang ingin memeriksanya: results.zip

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Threading.Tasks;

namespace FloodPaintAI
{
    class Node
    {   
        public byte Value;             //1-6
        public int Index;              //unique identifier, used for easily deepcopying the graph
        public List<Node> Neighbours;  
        public List<Tuple<int, int>> NeighboursPositions; //used by BuildGraph() 

        public int Depth;    //used by GetSumDistances() 
        public bool Checked; // 

        public Node(byte value, int index)
        {
            Value = value;      
            Index = index;          
        }

        public Node(Node node)
        {           
            Value = node.Value; 
            Index = node.Index;                     
        }
    }

    class Board
    {
        private const int SIZE = 19;
        private const int STARTPOSITION = 9;

        public Node Root;         //root of graph. each node is a set of contiguous, same color square
        public List<Node> Nodes;  //all nodes in the graph, used for deep copying


        public int EstimatedCost; //estimated cost, used by A* Pathfinding
        public List<byte> Solution;

        public Board(StreamReader input)
        {                   
            byte[,] board = new byte[SIZE, SIZE];
            for(int j = 0 ; j < SIZE ; j++)
            {
                string line = input.ReadLine();
                for(int i = 0 ; i < SIZE ; i++)         
                {                                       
                    board[j, i] = byte.Parse(line[i].ToString());
                }               
            }
            Solution = new List<byte>();
            BuildGraph(board);  
        }

        public Board(Board boardToCopy)
        {               
            //copy the graph            
            Nodes = new List<Node>(boardToCopy.Nodes.Count);
            foreach(Node nodeToCopy in boardToCopy.Nodes)
            {
                Node node = new Node(nodeToCopy);
                Nodes.Add(node);
            }

            //copy "Neighbours" property
            for(int i = 0 ; i < boardToCopy.Nodes.Count ; i++)
            {
                Node node = Nodes[i];
                Node nodeToCopy = boardToCopy.Nodes[i];

                node.Neighbours = new List<Node>(nodeToCopy.Neighbours.Count);
                foreach(Node neighbour in nodeToCopy.Neighbours)
                {
                    node.Neighbours.Add(Nodes[neighbour.Index]);
                }
            }

            Root = Nodes[boardToCopy.Root.Index];
            EstimatedCost = boardToCopy.EstimatedCost;          
            Solution = new List<byte>(boardToCopy.Solution);            
        }

        private void BuildGraph(byte[,] board)
        {                       
            int[,] nodeIndexes = new int[SIZE, SIZE];
            Nodes = new List<Node>();

            //check how much sets we have (1st pass)
            for(int j = 0 ; j < SIZE ; j++)
            {
                for(int i = 0 ; i < SIZE ; i++)         
                {               
                    if(nodeIndexes[j, i] == 0) //not already visited                    
                    {
                        Node newNode = new Node(board[j, i], Nodes.Count);                      
                        newNode.NeighboursPositions = new List<Tuple<int, int>>();
                        Nodes.Add(newNode);

                        BuildGraphFloodFill(board, nodeIndexes, newNode, i, j, board[j, i]);
                    }
                }       
            }

            //set neighbours and root (2nd pass)
            foreach(Node node in Nodes)
            {
                node.Neighbours = new List<Node>();
                node.Neighbours.AddRange(node.NeighboursPositions.Select(x => nodeIndexes[x.Item2, x.Item1]).Distinct().Select(x => Nodes[x - 1]));
                node.NeighboursPositions = null;                
            }
            Root = Nodes[nodeIndexes[STARTPOSITION, STARTPOSITION] - 1];            
        }

        private void BuildGraphFloodFill(byte[,] board, int[,] nodeIndexes, Node node, int startx, int starty, byte floodvalue)
        {
            Queue<Tuple<int, int>> queue = new Queue<Tuple<int, int>>();
            queue.Enqueue(new Tuple<int, int>(startx, starty));

            while(queue.Count > 0)
            {
                Tuple<int, int> position = queue.Dequeue();
                int x = position.Item1;
                int y = position.Item2;

                if(x >= 0 && x < SIZE && y >= 0 && y < SIZE)
                {
                    if(nodeIndexes[y, x] == 0 && board[y, x] == floodvalue)
                    {
                        nodeIndexes[y, x] = node.Index + 1;

                        queue.Enqueue(new Tuple<int, int>(x + 1, y));
                        queue.Enqueue(new Tuple<int, int>(x - 1, y));
                        queue.Enqueue(new Tuple<int, int>(x, y + 1));
                        queue.Enqueue(new Tuple<int, int>(x, y - 1));                                           
                    }               
                    if(board[y, x] != floodvalue)
                        node.NeighboursPositions.Add(position);                         
                }       
            }
        }

        public int GetEstimatedCost()
        {       
            Board current = this;

            //copy current board and play the best color until the end.
            //number of moves required to go the end is the heuristic
            //estimated cost = current cost + heuristic
            while(!current.IsSolved())
            {
                int minSumDistance = int.MaxValue;
                Board minBoard = null;

                //find color which give the minimum sum of distance from root to each other node
                foreach(byte i in current.Root.Neighbours.Select(x => x.Value).Distinct())
                {
                    Board copy = new Board(current);
                    copy.Play(i);                   

                    int distance = copy.GetSumDistances();                  

                    if(distance < minSumDistance)
                    {
                        minSumDistance = distance;
                        minBoard = copy;
                    }
                }
                current = minBoard;
            }           
            return current.Solution.Count;
        }

        public int GetSumDistances()
        {
            //get sum of distances from root to each other node, using BFS
            int sumDistances = 0;           

            //reset marker
            foreach(Node n in Nodes)
            {
                n.Checked = false;                                  
            }

            Queue<Node> queue = new Queue<Node>();
            Root.Checked = true;
            Root.Depth = 0; 
            queue.Enqueue(Root);

            while(queue.Count > 0)
            {
                Node current = queue.Dequeue();                             
                foreach(Node n in current.Neighbours)
                {
                    if(!n.Checked)          
                    {                                   
                        n.Checked = true;                                               
                        n.Depth = current.Depth + 1;
                        sumDistances += n.Depth;            
                        queue.Enqueue(n);   
                    }               
                }
            }
            return sumDistances;
        }       

        public void Play(byte value)            
        {
            //merge root node with other neighbours nodes, if color is matching
            Root.Value = value;
            List<Node> neighboursToRemove = Root.Neighbours.Where(x => x.Value == value).ToList();
            List<Node> neighboursToAdd = neighboursToRemove.SelectMany(x => x.Neighbours).Except((new Node[] { Root }).Concat(Root.Neighbours)).ToList();

            foreach(Node n in neighboursToRemove)
            {
                foreach(Node m in n.Neighbours)
                {
                    m.Neighbours.Remove(n);
                }
                Nodes.Remove(n);
            }   

            foreach(Node n in neighboursToAdd)
            {
                Root.Neighbours.Add(n);         
                n.Neighbours.Add(Root); 
            }           

            //re-synchronize node index
            for(int i = 0 ; i < Nodes.Count ; i++)
            {
                Nodes[i].Index = i;
            }           
            Solution.Add(value);
        }

        public bool IsSolved()
        {           
            //return Nodes.Count == 1;
            return Root.Neighbours.Count == 0;  
        }           
    }


    class Program
    {       
        public static List<byte> Solve(Board input)
        {
            //A* Pathfinding            
            LinkedList<Board> open = new LinkedList<Board>();       
            input.EstimatedCost = input.GetEstimatedCost();
            open.AddLast(input);            

            while(open.Count > 0)
            {                       
                Board current = open.First.Value;
                open.RemoveFirst();

                if(current.IsSolved())
                {
                    return current.Solution;                
                }
                else
                {
                    //play all neighbours nodes colors
                    foreach(byte i in current.Root.Neighbours.Select(x => x.Value).Distinct())
                    {                       
                        Board newBoard = new Board(current);
                        newBoard.Play(i);           
                        newBoard.EstimatedCost = newBoard.GetEstimatedCost();   

                        //insert board to open list
                        bool inserted = false;
                        for(LinkedListNode<Board> node = open.First ; node != null ; node = node.Next)
                        {                               
                            if(node.Value.EstimatedCost > newBoard.EstimatedCost)
                            {
                                open.AddBefore(node, newBoard);
                                inserted = true;
                                break;
                            }
                        }       
                        if(!inserted)
                            open.AddLast(newBoard);                                                 
                    }   
                }   
            }
            throw new Exception(); //no solution found, impossible
        }   

        public static void Main(string[] args)
        {                   
            using (StreamReader sr = new StreamReader("floodtest"))
            {   
                while(!sr.EndOfStream)
                {                               
                    List<Board> boards = new List<Board>();
                    while(!sr.EndOfStream && boards.Count < 100)
                    {
                        Board board = new Board(sr);                        
                        sr.ReadLine(); //skip empty line
                        boards.Add(board);
                    }                                           
                    List<byte>[] solutions = new List<byte>[boards.Count];                                          
                    Parallel.For(0, boards.Count, i => 
                    {                               
                        solutions[i] = Solve(boards[i]); 
                    });                                         
                    foreach(List<byte> solution in solutions)
                    {
                        Console.WriteLine(string.Join(string.Empty, solution));                                             
                    }       
                }               
            }
        }
    }
}

Lebih detail tentang cara kerjanya:

Ini menggunakan algoritma A * Pathfinding .

Yang sulit adalah menemukan yang baik heuristic. Jika heuristicitu meremehkan biaya, itu akan tampil seperti hampir seperti algoritma Dijkstra dan karena kompleksitas papan 19x19 dengan 6 warna itu akan berjalan selamanya. Jika itu melebih-lebihkan biayanya, ia akan menyatu dengan cepat ke sebuah solusi tetapi tidak akan memberikan yang baik sama sekali (sesuatu seperti 26 gerakan adalah 19 adalah mungkin). Menemukan yang sempurnaheuristic yang memberikan jumlah langkah yang tepat untuk mencapai solusi adalah yang terbaik (akan cepat dan akan memberikan solusi sebaik mungkin). Tidak mungkin (kecuali saya salah). Sebenarnya perlu menyelesaikan papan itu sendiri terlebih dahulu, sementara ini yang sebenarnya Anda coba lakukan (masalah ayam dan telur)

Saya mencoba banyak hal, inilah yang bekerja untuk heuristic:

• Saya membuat grafik papan saat ini untuk mengevaluasi. Masing-masing nodemewakili satu set kotak, berdekatan berwarna sama. Menggunakan itugraph , saya dapat dengan mudah menghitung jarak minimal yang tepat dari pusat ke node lain. Misalnya jarak dari tengah ke kiri atas adalah 10, karena minimal 10 warna memisahkannya.
• Untuk menghitung heuristic: Saya memainkan papan saat ini sampai akhir. Untuk setiap langkah, saya memilih warna yang akan meminimalkan jumlah jarak dari root ke semua node lainnya.

• Jumlah gerakan yang diperlukan untuk mencapai tujuan tersebut adalah heuristic.

• Estimated cost(digunakan oleh A *) = moves so far+heuristic

Ini tidak sempurna karena sedikit melebih-lebihkan biaya (sehingga solusi yang tidak optimal ditemukan). Pokoknya itu cepat untuk menghitung dan memberikan hasil yang baik.

Saya bisa mendapatkan peningkatan kecepatan besar dengan menggunakan grafik untuk melakukan semua operasi. Pada awalnya saya punya 2-dimensionarray. Saya membanjiri dan menghitung ulang grafik bila diperlukan (misalnya: untuk menghitung heuristik). Sekarang semuanya dilakukan dengan menggunakan grafik, yang dihitung hanya di awal. Untuk mensimulasikan langkah-langkah, mengisi banjir tidak diperlukan lagi, saya menggabungkan node sebagai gantinya. Ini jauh lebih cepat.

Python - 10.800.000 langkah

Sebagai solusi referensi tempat terakhir, pertimbangkan urutan ini:

print "123456" * 18

Bersepeda melalui semua waktu warna nberarti setiap nlangkah persegi akan dijamin dengan warna yang sama dengan pusat persegi. Setiap kotak paling banyak 18 langkah dari pusat, jadi 18 siklus akan menjamin semua kotak memiliki warna yang sama. Kemungkinan besar itu akan selesai dalam waktu kurang dari itu, tetapi program tidak diharuskan untuk berhenti begitu semua kotak berwarna sama; itu jauh lebih bermanfaat untuk melakukannya. Prosedur konstan ini adalah 108 langkah per test case, dengan total 10.800.000.

Jawa - 2.480.714 langkah

Saya membuat sedikit kesalahan sebelumnya (saya meletakkan satu kalimat penting sebelum satu lingkaran, bukan dalam satu lingkaran:

import java.io.*;

public class HerjanPaintAI {

    BufferedReader r;
    String[] map = new String[19];
    char[][] colors = new char[19][19];
    boolean[][] reached = new boolean[19][19], checked = new boolean[19][19];
    int[] colorCounter = new int[6];
    String answer = "";
    int mapCount = 0, moveCount = 0;

    public HerjanPaintAI(){
        nextMap();

        while(true){

            int bestMove = nextRound();
            answer += bestMove;
            char t = Character.forDigit(bestMove, 10);
            for(int x = 0; x < 19; x++){
                for(int y = 0; y < 19; y++){
                    if(reached[x][y]){
                        colors[x][y] = t;
                    }else if(checked[x][y]){
                        if(colors[x][y] == t){
                            reached[x][y] = true;
                        }
                    }
                }
            }

            boolean gameOver = true;
            for(int x = 0; x < 19; x++){
                for(int y = 0; y < 19; y++){
                    if(!reached[x][y]){
                        gameOver = false;
                        break;
                    }
                }
            }

            for(int x = 0; x < 19; x++){
                for(int y = 0; y < 19; y++){
                    checked[x][y] = false;
                }
            }
            for(int i = 0; i < 6; i++)
                colorCounter[i] = 0;

            if(gameOver)
                nextMap();
        }
    }

    int nextRound(){
        for(int x = 0; x < 19; x++){
            for(int y = 0; y < 19; y++){
                if(reached[x][y]){//check what numbers are next to the reached numbers...
                    check(x, y);
                }
            }
        }

        int[] totalColorCount = new int[6];
        for(int x = 0; x < 19; x++){
            for(int y = 0; y < 19; y++){
                totalColorCount[Character.getNumericValue(colors[x][y])-1]++;
            }
        }

        for(int i = 0; i < 6; i++){
            if(totalColorCount[i] != 0 && totalColorCount[i] == colorCounter[i]){//all of this color can be reached
                return i+1;
            }
        }

        int index = -1, number = 0;
        for(int i = 0; i < 6; i++){
            if(colorCounter[i] > number){
                index = i;
                number = colorCounter[i];
            }
        }

        return index+1;
    }

    void check(int x, int y){
        if(x<18)
            handle(x+1, y, x, y);
        if(x>0)
            handle(x-1, y, x, y);
        if(y<18)
            handle(x, y+1, x, y);
        if(y>0)
            handle(x, y-1, x, y);
    }

    void handle(int x2, int y2, int x, int y){
        if(!reached[x2][y2] && !checked[x2][y2]){
            checked[x2][y2] = true;
            if(colors[x2][y2] == colors[x][y]){
                reached[x2][y2] = true;
                check(x2, y2);
            }else{
                colorCounter[Character.getNumericValue(colors[x2][y2])-1]++;
                checkAround(x2, y2);
            }
        }
    }

    void checkAround(int x2, int y2){
        if(x2<18)
            handleAround(x2+1, y2, x2, y2);
        if(x2>0)
            handleAround(x2-1, y2, x2, y2);
        if(y2<18)
            handleAround(x2, y2+1, x2, y2);
        if(y2>0)
            handleAround(x2, y2-1, x2, y2);
    }

    void handleAround(int x2, int y2, int x, int y){
        if(!reached[x2][y2] && !checked[x2][y2]){
            if(colors[x2][y2] == colors[x][y]){
                checked[x2][y2] = true;
                colorCounter[Character.getNumericValue(colors[x2][y2])-1]++;
                checkAround(x2, y2);
            }
        }
    }

    void nextMap(){
        moveCount += answer.length();
        System.out.println(answer);
        answer = "";

        for(int x = 0; x < 19; x++){
            for(int y = 0; y < 19; y++){
                reached[x][y] = false;
            }
        }

        reached[9][9] = true;

        try {
            if(r == null)
                r = new BufferedReader(new FileReader("floodtest"));

            for(int i = 0; i < 19; i++){
                map[i] = r.readLine();
            }
            r.readLine();//empty line

            if(map[0] == null){
                System.out.println("Maps solved: " + mapCount + " Steps: " + moveCount);
                r.close();
                System.exit(0);
            }
        } catch (Exception e) {e.printStackTrace();}

        mapCount++;

        for(int x = 0; x < 19; x++){
            colors[x] = map[x].toCharArray();
        }
    }

    public static void main(String[] a){
        new HerjanPaintAI();
    }
}

C - 2,075,452

Saya tahu saya sangat terlambat ke pesta, tetapi saya melihat tantangan ini dan ingin mencobanya.

#include <assert.h>
#include <math.h>
#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>

uint64_t rand_state;

uint64_t rand_u64(void) {
    return (rand_state = rand_state * 6364136223846793005ULL + 1442695040888963407ULL);
}

uint64_t better_rand_u64(void) {
    uint64_t r = rand_u64();
    r ^= ((r >> 32) >> (r >> 60));
    return r + 1442695040888963407ULL;
}

uint32_t rand_u32(void) {return rand_u64() >> 32;}

float normal(float mu, float sigma) {
    uint64_t t = 0;
    for (int i = 0; i < 6; i++) {
        uint64_t r = rand_u64();
        uint32_t a = r;
        uint32_t b = r >> 32;
        t += a;
        t += b;
    }
    return ((float)t / (float)UINT32_MAX - 6) * sigma + mu;
}

typedef struct {
    uint8_t x;
    uint8_t y;
} Position;

#define ncolors 6
#define len 19
#define cells (len * len)
#define max_steps (len * (ncolors - 1))
#define center_x 9
#define center_y 9
#define center ((Position){center_x, center_y})

uint64_t zobrist_table[len][len];

void init_zobrist() {
    for (int y = 0; y < len; y++) {
        for (int x = 0; x < len; x++) {
            zobrist_table[y][x] = better_rand_u64();
        }
    }
}

typedef struct {
    uint64_t hash;
    uint8_t grid[len][len];
    bool interior[len][len];
    int boundary_size;
    Position boundary[cells];
} Grid;


void transition(Grid* grid, uint8_t color, int* surrounding_counts) {
    int i = 0;
    while (i < grid->boundary_size) {
        Position p = grid->boundary[i];
        uint8_t x = p.x;
        uint8_t y = p.y;
        bool still_boundary = false;
        for (int dx = -1; dx <= 1; dx++) {
            for (int dy = -1; dy <= 1; dy++) {
                if (!(dx == 0 || dy == 0)) {
                    continue;
                }
                int8_t x1 = x + dx;
                if (!(0 <= x1 && x1 < len)) {
                    continue;
                }
                int8_t y1 = y + dy;
                if (!(0 <= y1 && y1 < len)) {
                    continue;
                }
                if (grid->interior[y1][x1]) {
                    continue;
                }
                uint8_t color1 = grid->grid[y1][x1];
                if (color1 == color) {
                    grid->boundary[grid->boundary_size++] = (Position){x1, y1};
                    grid->interior[y1][x1] = true;
                    grid->hash ^= zobrist_table[y1][x1];
                } else {
                    surrounding_counts[color1]++;
                    still_boundary = true;
                }
            }
        }
        if (still_boundary) {
            i += 1;
        } else {
            grid->boundary[i] = grid->boundary[--grid->boundary_size]; 
        }
    }
}

void reset_grid(Grid* grid, int* surrounding_counts) {
    grid->hash = 0;
    memset(surrounding_counts, 0, ncolors * sizeof(int)); 
    memset(&grid->interior, 0, sizeof(grid->interior));
    grid->interior[center_y][center_x] = true;
    grid->boundary_size = 0;
    grid->boundary[grid->boundary_size++] = center; 
    transition(grid, grid->grid[center_y][center_x], surrounding_counts);
}

bool load_grid(FILE* fp, Grid* grid) {
    memset(grid, 0, sizeof(*grid));
    char buf[19 + 2];
    size_t row = 0;
    while ((fgets(buf, sizeof(buf), fp)) && row < 19) {
        if (strlen(buf) != 20) {
            break;
        }
        for (int i = 0; i < 19; i++) {
            if (!('1' <= buf[i] && buf[i] <= '6')) {
                return false;
            }
            grid->grid[row][i] = buf[i] - '1';
        }
        row++;
    }
    return row == 19;
}

typedef struct Node Node;

struct Node {
    uint64_t hash;
    float visit_counts[ncolors];
    float mean_cost[ncolors];
    float sse[ncolors];
};

#define iters 15000
#define pool_size 32768
#define pool_nodes (pool_size + 100)
#define pool_mask (pool_size - 1)

Node pool[pool_nodes];

void init_node(Node* node, uint64_t hash, int* surrounding_counts) {
    node->hash = hash;
    for (int i = 0; i < ncolors; i++) {
        if (surrounding_counts[i]) {
            node->visit_counts[i] = 1;
            node->mean_cost[i] = 20;
            node->sse[i] = 400;
        }
    }
}

Node* lookup_node(uint64_t hash) {
    size_t index = hash & pool_mask;
    for (int i = index;; i++) {
        uint64_t h = pool[i].hash;
        if (h == hash || !h) {
            return pool + i;
        }
    }
}

int rollout(Grid* grid, int* surrounding_counts, char* solution) {
    for (int i = 0;; i++) {
        int nonzero = 0;
        uint8_t colors[6];
        for (int i = 0; i < ncolors; i++) {
            if (surrounding_counts[i]) {
                colors[nonzero++] = i;
            }
        }
        if (!nonzero) {
            return i;
        }
        uint8_t color = colors[rand_u32() % nonzero]; 
        *(solution++) = color;
        assert(grid->boundary_size);
        memset(surrounding_counts, 0, 6 * sizeof(int));
        transition(grid, color, surrounding_counts);
    }
}

int simulate(Node* node, Grid* grid, int depth, char* solution) {
    float best_cost = INFINITY;
    uint8_t best_color = 255;
    for (int color = 0; color < ncolors; color++) {
        float n = node->visit_counts[color];
        if (node->visit_counts[color] == 0) {
            continue;
        }
        float sigma = sqrt(node->sse[color] / (n * n));
        float cost = node->mean_cost[color];
        cost = normal(cost, sigma);
        if (cost < best_cost) {
            best_color = color;
            best_cost = cost;
        }
    }
    if (best_color == 255) {
        return 0;
    }
    *solution++ = best_color;
    int score;
    int surrounding_counts[ncolors] = {0};
    transition(grid, best_color, surrounding_counts);
    Node* child = lookup_node(grid->hash);
    if (!child->hash) {
        init_node(child, grid->hash, surrounding_counts);
        score = rollout(grid, surrounding_counts, solution);
    } else {
        score = simulate(child, grid, depth + 1, solution);
    }
    score++;
    float n1 = ++node->visit_counts[best_color];
    float u0 = node->mean_cost[best_color];
    float u1 = node->mean_cost[best_color] = u0 + (score - u0) / n1;
    node->sse[best_color] += (score - u0) * (score - u1);
    return score;
}

int main(void) {
    FILE* fp;
    if (!(fp = fopen("floodtest", "r"))) {
        return 1;
    }
    Grid grid;
    init_zobrist();
    while (load_grid(fp, &grid)) {

        memset(pool, 0, sizeof(pool));
        int surrounding_counts[ncolors] = {0};

        reset_grid(&grid, surrounding_counts);
        Node root = {0};

        init_node(&root, grid.hash, surrounding_counts);

        char solution[max_steps] = {0};
        char best_solution[max_steps] = {0};

        int min_score = INT_MAX;

        uint64_t prev_hash = 0;
        uint64_t hash = 0;
        int same_count = 0;

        for (int iter = 0; iter < iters; iter++) {
            reset_grid(&grid, surrounding_counts);
            int score = simulate(&root, &grid, 1, solution);
            if (score < min_score) {
                min_score = score;
                memcpy(best_solution, solution, score);
            }
            hash = 0;
            for (int i = 0; i < score; i++) {
                hash ^= zobrist_table[i%len][(int)solution[i]];
            }
            if (hash == prev_hash) {
                same_count++;
                if (same_count >= 10) {
                    break;
                }
            } else {
                same_count = 0;
                prev_hash = hash;
            }
        }
        int i;
        for (i = 0; i < min_score; i++) {
            best_solution[i] += '1';
        }
        best_solution[i++] = '\n';
        best_solution[i++] = '\0';
        printf(best_solution);
        fflush(stdout);
    }
    return 0;
}

Algoritma ini didasarkan pada Pencarian Pohon Monte-Carlo dengan pengambilan sampel Thompson, dan tabel transposisi untuk mengurangi ruang pencarian. Butuh sekitar 12 jam di mesin saya. Jika Anda ingin memeriksa hasilnya, Anda dapat menemukannya di https://dropfile.to/pvjYDMV .

Jawa - 2.434.108 2.588.847 langkah

Saat ini menang (~ 46K di atas Herjan) pada 4/26

Welp, jadi bukan hanya MrBackend mengalahkan saya, tetapi saya menemukan bug yang menghasilkan skor yang sangat bagus. Sudah diperbaiki sekarang (juga di verifikasi! Ack), tapi sayangnya saya tidak punya waktu saat ini untuk mencoba dan mengambil kembali mahkota. Akan berusaha nanti.

Ini didasarkan pada solusi saya yang lain, tetapi alih-alih mengecat dengan warna yang paling umum pada tepi isian, cat ini melukis dengan warna yang akan menghasilkan pinggiran yang memiliki banyak kotak yang berdekatan dengan warna yang sama. Sebut saja LookAheadPainter. Saya akan golf nanti jika perlu.

import java.io.*;
import java.util.*;

public class LookAheadPainter {

    static final boolean PRINT_FULL_OUTPUT = true;

    public static void main(String[] a) throws IOException {

        int totalSteps = 0, numSolved = 0;

        char[] board = new char[361];
        Scanner s = new Scanner(new File("floodtest"));
        long startTime = System.nanoTime();

        caseloop: while (s.hasNextLine()) {
            for (int l = 0; l < 19; l++) {
                String line = s.nextLine();
                if (line.isEmpty())
                    continue caseloop;
                System.arraycopy(line.toCharArray(), 0, board, l * 19, 19);
            }

            List<Character> colorsUsed = new ArrayList<>();

            for (;;) {

                FillResult fill = new FillResult(board, board[180], (char) 48, null);

                if (fill.nodesFilled.size() == 361)
                    break;

                int[] branchSizes = new int[7];

                for (int i = 1; i < 7; i++) {
                    List<Integer> seeds = new ArrayList<>();
                    for (Integer seed : fill.edges)
                        if (board[seed] == i + 48)
                            seeds.add(seed);

                    branchSizes[i] = new FillResult(fill.filledBoard, (char) (i + 48), (char) 48, seeds).nodesFilled.size();
                }

                int maxSize = 0;
                char bestColor = 0;

                for (int i = 1; i < 7; i++)
                    if (branchSizes[i] > maxSize) {
                        maxSize = branchSizes[i];
                        bestColor = (char) (i + 48);
                    }

                for (int i : fill.nodesFilled)
                    board[i] = bestColor;

                colorsUsed.add(bestColor);
                totalSteps++;
            }
            numSolved++;

            if (PRINT_FULL_OUTPUT) {
                if (numSolved % 1000 == 0)
                    System.out.println("Solved: " + numSolved); // So you know it's working
                String out = "";
                for (Character c : colorsUsed)
                    out += c;
                System.out.println(out);
            }

        }
        s.close();
        System.out.println("\nTotal steps to solve all cases: " + totalSteps);
        System.out.printf("\nSolved %d test cases in %.2f seconds", numSolved, (System.nanoTime() - startTime) / 1000000000.);
    }

    private static class FillResult {

        Set<Integer> nodesFilled, edges;
        char[] filledBoard;

        FillResult(char[] board, char target, char replacement, List<Integer> seeds) {
            Stack<Integer> nodes = new Stack<>();
            nodesFilled = new HashSet<>();
            edges = new HashSet<>();

            if (seeds == null)
                nodes.push(180);
            else
                for (int i : seeds)
                    nodes.push(i);

            filledBoard = new char[361];
            System.arraycopy(board, 0, filledBoard, 0, 361);

            while (!nodes.empty()) {
                int n = nodes.pop();
                if (n < 0 || n > 360)
                    continue;
                if (filledBoard[n] == target) {
                    filledBoard[n] = replacement;
                    nodesFilled.add(n);
                    if (n % 19 > 0)
                        nodes.push(n - 1);
                    if (n % 19 < 18)
                        nodes.push(n + 1);
                    if (n / 19 > 0)
                        nodes.push(n - 19);
                    if (n / 19 < 18)
                        nodes.push(n + 19);
                } else
                    edges.add(n);
            }
        }
    }
}

EDIT: Saya menulis verifier, jangan ragu untuk menggunakannya, ia mengharapkan file steps.txt yang berisi langkah-langkah output program Anda dan juga file floodtest: Edit-Edit: (Lihat OP)

Jika ada yang menemukan masalah, tolong laporkan kepada saya!

C - 2.480.714 langkah

Masih belum optimal, tetapi sekarang lebih cepat dan skor lebih baik.

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

char map[19][19], reach[19][19];
int reachsum[6], totalsum[6];

bool loadmap(FILE *fp)
{
    char buf[19 + 2];
    size_t row = 0;

    while (fgets(buf, sizeof buf, fp) && row < 19) {
        if (strlen(buf) != 20)
            break;
        memcpy(map[row++], buf, 19);
    }
    return row == 19;
}

void calcreach(bool first, size_t row, size_t col);
void check(char c, bool first, size_t row, size_t col)
{
    if (map[row][col] == c)
        calcreach(first, row, col);
    else if (first)
        calcreach(false, row, col);
}

void calcreach(bool first, size_t row, size_t col)
{
    char c = map[row][col];

    reach[row][col] = c;
    reachsum[c - '1']++;
    if (row < 18 && !reach[row + 1][col])
        check(c, first, row + 1, col);
    if (col < 18 && !reach[row][col + 1])
        check(c, first, row, col + 1);
    if (row > 0 && !reach[row - 1][col])
        check(c, first, row - 1, col);
    if (col > 0 && !reach[row][col - 1])
        check(c, first, row, col - 1);
}

void calctotal()
{
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            totalsum[map[row][col] - '1']++;
}

void apply(char c)
{
    char d = map[9][9];
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            if (reach[row][col] == d)
                map[row][col] = c;
}

int main()
{
    char c, best;
    size_t steps = 0;
    FILE *fp;

    if (!(fp = fopen("floodtest", "r")))
        return 1;

    while (loadmap(fp)) {
        do {
            memset(reach, 0, sizeof reach);
            memset(reachsum, 0, sizeof reachsum);
            calcreach(true, 9, 9);
            if (reachsum[map[9][9] - '1'] == 361)
                break;

            memset(totalsum, 0, sizeof totalsum);
            calctotal();

            reachsum[map[9][9] - '1'] = 0;
            for (best = 0, c = 0; c < 6; c++) {
                if (!reachsum[c])
                    continue;
                if (reachsum[c] == totalsum[c]) {
                    best = c;
                    break;
                } else if (reachsum[c] > reachsum[best]) {
                    best = c;
                }
            }

            apply(best + '1');
        } while (++steps);
    }

    fclose(fp);

    printf("steps: %zu\n", steps);
    return 0;
}

Java - 2.245.529 2.201.995 langkah

Pencarian pohon paralel & caching pada kedalaman 5, meminimalkan jumlah "pulau". Karena peningkatan dari kedalaman 4 ke kedalaman 5 sangat kecil, saya tidak berpikir ada banyak gunanya memperbaikinya lebih jauh. Tetapi jika perlu perbaikan, firasat saya mengatakan untuk bekerja dengan menghitung jumlah pulau sebagai perbedaan antara dua negara, bukannya menghitung ulang semuanya.

Saat ini output ke stdout, sampai saya tahu format input verifier.

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.AbstractList;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.BitSet;
import java.util.Collection;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.RecursiveTask;
import java.util.concurrent.locks.ReadWriteLock;
import java.util.concurrent.locks.ReentrantReadWriteLock;

public class FloodPaint {

    private static final ForkJoinPool FORK_JOIN_POOL = new ForkJoinPool();

    public static void main(String[] arg) throws IOException, InterruptedException, ExecutionException {
        try (BufferedReader reader = new BufferedReader(new FileReader("floodtest"))) {
            int sum = 0;
            State initState = readNextInitState(reader);
            while (initState != null) {
                List<Integer> solution = generateSolution(initState);
                System.out.println(solution);
                sum += solution.size();
                initState = readNextInitState(reader);
            }
            System.out.println(sum);
        }
    }

    private static State readNextInitState(BufferedReader reader) throws IOException {
        int[] initGrid = new int[State.DIM * State.DIM];
        String line = reader.readLine();
        while ((line != null) && line.isEmpty()) {
            line = reader.readLine();
        }
        if (line == null) {
            return null;
        }
        for (int rowNo = 0; rowNo < State.DIM; ++rowNo) {
            for (int colNo = 0; colNo < State.DIM; ++colNo) {
                initGrid[(State.DIM * rowNo) + colNo] = line.charAt(colNo) - '0';
            }
            line = reader.readLine();
        }
        return new State(initGrid);
    }

    private static List<Integer> generateSolution(State initState) throws InterruptedException, ExecutionException {
        List<Integer> solution = new LinkedList<>();
        StateFactory stateFactory = new StateFactory();
        State state = initState;
        while (!state.isSolved()) {
            int num = findGoodNum(state, stateFactory);
            solution.add(num);
            state = state.getNextState(num, stateFactory);
        }
        return solution;
    }

    private static int findGoodNum(State state, StateFactory stateFactory) throws InterruptedException, ExecutionException {
        SolverTask task = new SolverTask(state, stateFactory);
        FORK_JOIN_POOL.invoke(task);
        return task.get();
    }

}

class SolverTask extends RecursiveTask<Integer> {

    private static final int DEPTH = 5;

    private final State state;
    private final StateFactory stateFactory;

    SolverTask(State state, StateFactory stateFactory) {
        this.state = state;
        this.stateFactory = stateFactory;
    }

    @Override
    protected Integer compute() {
        try {
            Map<Integer,AnalyzerTask> tasks = new HashMap<>();
            for (int num = 1; num <= 6; ++num) {
                if (num != state.getCenterNum()) {
                    State nextState = state.getNextState(num, stateFactory);
                    AnalyzerTask task = new AnalyzerTask(nextState, DEPTH - 1, stateFactory);
                    tasks.put(num, task);
                }
            }
            invokeAll(tasks.values());
            int bestValue = Integer.MAX_VALUE;
            int bestNum = -1;
            for (Map.Entry<Integer,AnalyzerTask> taskEntry : tasks.entrySet()) {
                int value = taskEntry.getValue().get();
                if (value < bestValue) {
                    bestValue = value;
                    bestNum = taskEntry.getKey();
                }
            }
            return bestNum;
        } catch (InterruptedException | ExecutionException ex) {
            throw new RuntimeException(ex);
        }
    }

}

class AnalyzerTask extends RecursiveTask<Integer> {

    private static final int DEPTH_THRESHOLD = 3;

    private final State state;
    private final int depth;
    private final StateFactory stateFactory;

    AnalyzerTask(State state, int depth, StateFactory stateFactory) {
        this.state = state;
        this.depth = depth;
        this.stateFactory = stateFactory;
    }

    @Override
    protected Integer compute() {
        return (depth < DEPTH_THRESHOLD) ? analyze() : split();
    }

    private int analyze() {
        return analyze(state, depth);
    }

    private int analyze(State state, int depth) {
        if (state.isSolved()) {
            return -depth;
        }
        if (depth == 0) {
            return state.getNumIslands();
        }
        int bestValue = Integer.MAX_VALUE;
        for (int num = 1; num <= 6; ++num) {
            if (num != state.getCenterNum()) {
                State nextState = state.getNextState(num, stateFactory);
                int nextValue = analyze(nextState, depth - 1);
                bestValue = Math.min(bestValue, nextValue);
            }
        }
        return bestValue;
    }

    private int split() {
        try {
            if (state.isSolved()) {
                return -depth;
            }
            Collection<AnalyzerTask> tasks = new ArrayList<>(5);
            for (int num = 1; num <= 6; ++num) {
                State nextState = state.getNextState(num, stateFactory);
                AnalyzerTask task = new AnalyzerTask(nextState, depth - 1, stateFactory);
                tasks.add(task);
            }
            invokeAll(tasks);
            int bestValue = Integer.MAX_VALUE;
            for (AnalyzerTask task : tasks) {
                int nextValue = task.get();
                bestValue = Math.min(bestValue, nextValue);
            }
            return bestValue;
        } catch (InterruptedException | ExecutionException ex) {
            throw new RuntimeException(ex);
        }
    }

}

class StateFactory {

    private static final int INIT_CAPACITY = 40000;
    private static final float LOAD_FACTOR = 0.9f;

    private final ReadWriteLock cacheLock = new ReentrantReadWriteLock();
    private final Map<List<Integer>,State> cache = new HashMap<>(INIT_CAPACITY, LOAD_FACTOR);

    State get(int[] grid) {
        List<Integer> stateKey = new IntList(grid);
        State state;
        cacheLock.readLock().lock();
        try {
            state = cache.get(stateKey);
        } finally {
            cacheLock.readLock().unlock();
        }
        if (state == null) {
            cacheLock.writeLock().lock();
            try {
                state = cache.get(stateKey);
                if (state == null) {
                    state = new State(grid);
                    cache.put(stateKey, state);
                }
            } finally {
                cacheLock.writeLock().unlock();
            }
        }
        return state;
    }

}

class State {

    static final int DIM = 19;
    private static final int CENTER_INDEX = ((DIM * DIM) - 1) / 2;

    private final int[] grid;
    private int numIslands;

    State(int[] grid) {
        this.grid = grid;
        numIslands = calcNumIslands(grid);
    }

    private static int calcNumIslands(int[] grid) {
        int numIslands = 0;
        BitSet uncounted = new BitSet(DIM * DIM);
        uncounted.set(0, DIM * DIM);
        int index = -1;
        while (!uncounted.isEmpty()) {
            index = uncounted.nextSetBit(index + 1);
            BitSet island = new BitSet(DIM * DIM);
            generateIsland(grid, index, grid[index], island);
            ++numIslands;
            uncounted.andNot(island);
        }
        return numIslands;
    }

    private static void generateIsland(int[] grid, int index, int num, BitSet island) {
        if ((grid[index] == num) && !island.get(index)) {
            island.set(index);
            if ((index % DIM) > 0) {
                generateIsland(grid, index - 1, num, island);
            }
            if ((index % DIM) < (DIM - 1)) {
                generateIsland(grid, index + 1, num, island);
            }
            if ((index / DIM) > 0) {
                generateIsland(grid, index - DIM, num, island);
            }
            if ((index / DIM) < (DIM - 1)) {
                generateIsland(grid, index + DIM, num, island);
            }
        }
    }

    int getCenterNum() {
        return grid[CENTER_INDEX];
    }

    boolean isSolved() {
        return numIslands == 1;
    }

    int getNumIslands() {
        return numIslands;
    }

    State getNextState(int num, StateFactory stateFactory) {
        int[] nextGrid = grid.clone();
        if (num != getCenterNum()) {
            flood(nextGrid, CENTER_INDEX, getCenterNum(), num);
        }
        State nextState = stateFactory.get(nextGrid);
        return nextState;
    }

    private static void flood(int[] grid, int index, int fromNum, int toNum) {
        if (grid[index] == fromNum) {
            grid[index] = toNum;
            if ((index % 19) > 0) {
                flood(grid, index - 1, fromNum, toNum);
            }
            if ((index % 19) < (DIM - 1)) {
                flood(grid, index + 1, fromNum, toNum);
            }
            if ((index / 19) > 0) {
                flood(grid, index - DIM, fromNum, toNum);
            }
            if ((index / 19) < (DIM - 1)) {
                flood(grid, index + DIM, fromNum, toNum);
            }
        }
    }

}

class IntList extends AbstractList<Integer> implements List<Integer> {

    private final int[] arr;
    private int hashCode = -1;

    IntList(int[] arr) {
        this.arr = arr;
    }

    @Override
    public int size() {
        return arr.length;
    }

    @Override
    public Integer get(int index) {
        return arr[index];
    }

    @Override
    public Integer set(int index, Integer value) {
        int oldValue = arr[index];
        arr[index] = value;
        return oldValue;
    }

    @Override
    public boolean equals(Object obj) {
        if (this == obj) {
            return true;
        }
        if (obj instanceof IntList) {
            IntList arg = (IntList) obj;
            return Arrays.equals(arr, arg.arr);
        }
        return super.equals(obj);
    }

    @Override
    public int hashCode() {
        if (hashCode == -1) {
            hashCode = 1;
            for (int elem : arr) {
                hashCode = 31 * hashCode + elem;
            }
        }
        return hashCode;
    }

}

Entri Terakhir Saya: C - 2.384.020 langkah

Kali ini yang 'periksa semua kemungkinan' ... Skor ini diperoleh dengan Kedalaman yang ditetapkan pada 3. Kedalaman pada 5 harus memberikan ~ langkah 2.1M ... TERLALU LAMBAT. Kedalaman 20+ memberikan jumlah langkah seminimal mungkin (hanya memeriksa semua pertandingan dan kemenangan tersingkat dari kursus) ... Ini memiliki jumlah langkah paling sedikit, meskipun saya benci karena ini hanya sedikit lebih baik, tetapi kinerjanya payah. Saya lebih suka entri C saya yang lain, yang juga ada di posting ini.

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

char map[19][19], reach[19][19];
int reachsum[6], totalsum[6], mapCount = 0;
FILE *stepfile;

bool loadmap(FILE *fp)
{
    fprintf(stepfile, "%s", "\n");

    mapCount++;

    char buf[19 + 2];
    size_t row = 0;

    while (fgets(buf, sizeof buf, fp) && row < 19) {
        if (strlen(buf) != 20)
            break;
        memcpy(map[row++], buf, 19);
    }
    return row == 19;
}

void calcreach(bool first, size_t row, size_t col);
void check(char c, bool first, size_t row, size_t col)
{
    if (map[row][col] == c)
        calcreach(first, row, col);
    else if (first)
        calcreach(false, row, col);
}

void calcreach(bool first, size_t row, size_t col)
{
    char c = map[row][col];

    reach[row][col] = c;
    reachsum[c - '1']++;
    if (row < 18 && !reach[row + 1][col])
        check(c, first, row + 1, col);
    if (col < 18 && !reach[row][col + 1])
        check(c, first, row, col + 1);
    if (row > 0 && !reach[row - 1][col])
        check(c, first, row - 1, col);
    if (col > 0 && !reach[row][col - 1])
        check(c, first, row, col - 1);
}

void calctotal()
{
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            totalsum[map[row][col] - '1']++;
}

void apply(char c)
{
    char d = map[9][9];
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            if (reach[row][col] == d)
                map[row][col] = c;
}

int pown(int x, int y){
    int p = 1;
    for(int i = 0; i < y; i++){
        p = p * x;
    }

    return p;
}

int main()
{
    size_t steps = 0;
    FILE *fp;

    if (!(fp = fopen("floodtest", "r")))
        return 1;
    if(!(stepfile = fopen("steps.txt", "w")))
        return 1;

    const int depth = 5;
    char possibilities[pown(6, depth)][depth];
    int t = 0;
    for(int a = 0; a < 6; a++){
        for(int b = 0; b < 6; b++){
            for(int c = 0; c < 6; c++){
                for(int d = 0; d < 6; d++){
                    for(int e = 0; e < 6; e++){
                        possibilities[t][0] = (char)(a + '1');
                        possibilities[t][1] = (char)(b + '1');
                        possibilities[t][2] = (char)(c + '1');
                        possibilities[t][3] = (char)(d + '1');
                        possibilities[t++][4] = (char)(e + '1');
                    }
                }
            }
        }
    }
    poes:
    while (loadmap(fp)) {
        do {
            char map2[19][19];
            memcpy(map2, map, sizeof(map));

            memset(reach, 0, sizeof reach);
            memset(reachsum, 0, sizeof reachsum);
            calcreach(true, 9, 9);

            int best = 0, index = 0, end = depth;
            for(int i = 0; i < pown(6, depth); i++){
                for(int d = 0; d < end; d++){

                    apply(possibilities[i][d]);

                    memset(reach, 0, sizeof reach);
                    memset(reachsum, 0, sizeof reachsum);
                    calcreach(true, 9, 9);

                    if(reachsum[map[9][9] - '1'] == 361 && d < end){
                        end = d+1;
                        index = i;
                        break;
                    }
                }
                if(end == depth && best < reachsum[map[9][9] - '1']){
                    best = reachsum[map[9][9] - '1'];
                    index = i;
                }

                memcpy(map, map2, sizeof(map2));
                memset(reach, 0, sizeof reach);
                memset(reachsum, 0, sizeof reachsum);
                calcreach(true, 9, 9);
            }

            for(int d = 0; d < end; d++){

                apply(possibilities[index][d]);

                memset(reach, 0, sizeof reach);
                memset(reachsum, 0, sizeof reachsum);
                calcreach(true, 9, 9);

                fprintf(stepfile, "%c", possibilities[index][d]);
                steps++;
            }
            if(reachsum[map[9][9] - '1'] == 361)
                goto poes;
        } while (1);
    }

    fclose(fp);
    fclose(stepfile);

    printf("steps: %zu\n", steps);
    return 0;
}

Peningkatan AI lainnya ditulis dalam langkah C - 2.445.761

Berdasarkan SteelTermite:

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

char map[19][19], reach[19][19];
int reachsum[6], totalsum[6], mapCount = 0;
FILE *stepfile;

bool loadmap(FILE *fp)
{
    fprintf(stepfile, "%s", "\n");

    if(mapCount % 1000 == 0)
        printf("mapCount = %d\n", mapCount);

    mapCount++;

    char buf[19 + 2];
    size_t row = 0;

    while (fgets(buf, sizeof buf, fp) && row < 19) {
        if (strlen(buf) != 20)
            break;
        memcpy(map[row++], buf, 19);
    }
    return row == 19;
}

void calcreach(bool first, size_t row, size_t col);
void check(char c, bool first, size_t row, size_t col)
{
    if (map[row][col] == c)
        calcreach(first, row, col);
    else if (first)
        calcreach(false, row, col);
}

void calcreach(bool first, size_t row, size_t col)
{
    char c = map[row][col];

    reach[row][col] = c;
    reachsum[c - '1']++;
    if (row < 18 && !reach[row + 1][col])
        check(c, first, row + 1, col);
    if (col < 18 && !reach[row][col + 1])
        check(c, first, row, col + 1);
    if (row > 0 && !reach[row - 1][col])
        check(c, first, row - 1, col);
    if (col > 0 && !reach[row][col - 1])
        check(c, first, row, col - 1);
}

void calctotal()
{
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            totalsum[map[row][col] - '1']++;
}

void apply(char c)
{
    char d = map[9][9];
    size_t row, col;

    for (row = 0; row < 19; row++)
        for (col = 0; col < 19; col++)
            if (reach[row][col] == d)
                map[row][col] = c;
}

int main()
{
    char c, best, answer;
    size_t steps = 0;
    FILE *fp;

    if (!(fp = fopen("floodtest", "r")))
        return 1;
    if(!(stepfile = fopen("steps.txt", "w")))
            return 1;

    while (loadmap(fp)) {
        do {
            memset(reach, 0, sizeof reach);
            memset(reachsum, 0, sizeof reachsum);
            calcreach(true, 9, 9);
            if (reachsum[map[9][9] - '1'] == 361)
                break;

            memset(totalsum, 0, sizeof totalsum);
            calctotal();

            reachsum[map[9][9] - '1'] = 0;
            for (best = 0, c = 0; c < 6; c++) {
                if (!reachsum[c])
                    continue;
                if (reachsum[c] == totalsum[c]) {
                    best = c;
                    goto outLoop;
                } else if (reachsum[c] > reachsum[best]) {
                    best = c;
                }
            }

            char map2[19][19];
            memcpy(map2, map, sizeof(map));

            int temp = best;
            for(c = 0; c < 6; c++){

                if(c != best){

                    apply(c + '1');

                    memset(reach, 0, sizeof reach);
                    memset(reachsum, 0, sizeof reachsum);
                    calcreach(true, 9, 9);
                    if (reachsum[best] == totalsum[best]) {

                        memcpy(map, map2, sizeof(map2));
                        memset(reach, 0, sizeof reach);
                        memset(reachsum, 0, sizeof reachsum);
                        calcreach(true, 9, 9);

                        if(temp == -1)
                            temp = c;
                        else if(reachsum[c] > reachsum[temp])
                            temp = c;
                    }

                    memcpy(map, map2, sizeof(map2));
                    memset(reach, 0, sizeof reach);
                    memset(reachsum, 0, sizeof reachsum);
                    calcreach(true, 9, 9);
                }
            }

outLoop:    answer = (char)(temp + '1');
            fprintf(stepfile, "%c", answer);
            apply(answer);
        } while (++steps);
    }

    fclose(fp);
    fclose(stepfile);

    printf("steps: %zu\n", steps);
    return 0;
}

Java - 2.610.797 4.780.841 langkah

(Isi-Bug diperbaiki, skor sekarang secara dramatis lebih buruk -_-)

Ini adalah pengajuan algoritma referensi dasar saya, cukup membuat histogram dari kotak di tepi area yang dicat, dan melukis dengan warna yang paling umum. Menjalankan 100k dalam beberapa menit.

Jelas tidak akan menang, tetapi tentu saja itu tidak bertahan lama. Saya mungkin akan membuat pengajuan lain untuk hal-hal pintar. Jangan ragu untuk menggunakan algoritma ini sebagai titik awal.

Hapus komentar pada baris komentar untuk output penuh. Default untuk mencetak # langkah yang diambil.

import java.io.*;
import java.util.*;

public class PainterAI {

    public static void main(String[] args) throws IOException {

        int totalSteps = 0, numSolved = 0;

        char[] board = new char[361];
        Scanner s = new Scanner(new File("floodtest"));
        long startTime = System.nanoTime();
        caseloop: while (s.hasNextLine()) {
            for (int l = 0; l < 19; l++) {
                String line = s.nextLine();
                if (line.isEmpty())
                    continue caseloop;
                System.arraycopy(line.toCharArray(), 0, board, l * 19, 19);
            }

            List<Character> colorsUsed = new ArrayList<>();
            Stack<Integer> nodes = new Stack<>();

            for (;; totalSteps++) {
                char p = board[180];
                int[] occurrences = new int[7];
                nodes.add(180);
                int numToPaint = 0;
                while (!nodes.empty()) {
                    int n = nodes.pop();
                    if (n < 0 || n > 360)
                        continue;
                    if (board[n] == p) {
                        board[n] = 48;
                        numToPaint++;
                        if (n % 19 > 0)
                            nodes.push(n - 1);
                        if(n%19<18)
                            nodes.push(n + 1);
                        if(n/19>0)
                            nodes.push(n - 19);
                        if(n/19<18)
                            nodes.push(n + 19);
                    } else
                        occurrences[board[n] - 48]++;
                }
                if (numToPaint == 361)
                    break;
                char mostFrequent = 0;
                int times = -1;
                for (int i = 1; i < 7; i++)
                    if (occurrences[i] > times) {
                        times = occurrences[i];
                        mostFrequent = (char) (i + 48);
                    }
                for (int i = 0; i < 361; i++)
                    if (board[i] == 48)
                        board[i] = mostFrequent;
                //colorsUsed.add(mostFrequent);
            }
            numSolved++;

            /*String out = "";
            for (Character c : colorsUsed)
                out += c;
            System.out.println(out); //print output*/
        }
        s.close();
        System.out.println("Total steps to solve all cases: " + totalSteps);
        System.out.printf("\nSolved %d test cases in %.2f seconds", numSolved, (System.nanoTime() - startTime) / 1000000000.);
    }
}

Golf hingga 860 karakter (tidak termasuk baris baru untuk format), tetapi bisa menyusut lebih banyak jika saya ingin mencoba:

import java.io.*;import java.util.*;class P{
public static void main(String[]a)throws Exception{int t=0;char[]b=new char[361];
Scanner s=new Scanner(new File("floodtest"));c:while(s.hasNextLine()){
for(int l=0;l<19;l++){String L=s.nextLine();if(L.isEmpty())continue c;
System.arraycopy(L.toCharArray(),0,b,l*19,19);}List<Character>u=new ArrayList<>();
Stack<Integer>q=new Stack<>();for(int[]o=new int[7];;t++){char p=b[180];q.add(180);
int m=0;while(!q.empty()){int n=q.pop();if(n<0|n>360)continue;if(b[n]==p){b[n]=48;m++;
if(n%19>0)q.add(n-1);if(n%19<18)q.add(n+1);if(n/19>0)q.add(n-19);if(n/19<18)
q.add(n+19);}else o[b[n]-48]++;}if(m==361)break;
char f=0;int h=0;for(int i=1;i<7;i++)if(o[i]>h){h=o[i];f=(char)(i+48);}
for(int i=0;i<361;i++)if(b[i]==48)b[i]=f;u.add(f);}String y="";for(char c:u)y+=c;
System.out.println(y);}s.close();System.out.println("Steps: "+t);}}

Haskell - 2.475.056 langkah

Algoritma mirip dengan yang disarankan oleh MrBackend dalam komentar. Perbedaannya adalah: sarannya menemukan jalur termurah ke kuadrat biaya tertinggi, tambang dengan rakus mengurangi eksentrisitas grafik di setiap langkah.

import Data.Array
import qualified Data.Map as M
import Data.Word
import Data.List
import Data.Maybe
import Data.Function (on)
import Data.Monoid
import Control.Arrow
import Control.Monad (liftM)
import System.IO
import System.Environment
import Control.Parallel.Strategies
import Control.DeepSeq

type Grid v = Array (Word8,Word8) v

main = do
  (ifn:_) <- getArgs
  hr <- openFile ifn ReadMode
  sp <- liftM parseFile $ hGetContents hr
  let (len,sol) = turns (map solve sp `using` parBuffer 3 (evalList rseq))
  putStrLn $ intercalate "\n" $ map (concatMap show) sol
  putStrLn $ "\n\nTotal turns: " ++ (show len)

turns :: [[a]] -> (Integer,[[a]])
turns l = rl' 0 l where
  rl' c [] = (c,[])
  rl' c (k:r) = let
   s = c + genericLength k
   (s',l') = s `seq` rl' s r
   in (s',k:l')

centrepoint :: Grid v -> (Word8,Word8)
centrepoint g = let
  ((x0,y0),(x1,y1)) = bounds g
  med l h = let t = l + h in t `div` 2 + t `mod` 2
  in (med x0 x1, med y0 y1)

neighbours :: Grid v -> (Word8,Word8) -> [(Word8,Word8)]
neighbours g (x,y) = filter
  (inRange $ bounds g)
  [(x,y+1),(x+1,y),(x,y-1),(x-1,y)]

areas :: Eq v => Grid v -> [[(Word8,Word8)]]
areas g = p $ indices g where
  p [] = []
  p (a:r) = f : p (r \\ f) where
    f = s g [a] []
s g [] _ = []
s g (h:o) v = let
  n = filter (((==) `on` (g !)) h) $ neighbours g h
  in h : s g ((n \\ (o ++ v)) ++ o) (h : v)

applyFill :: Eq v => v -> Grid v -> Grid v
applyFill c g = g // (zip fa $ repeat c) where
  fa = s g [centrepoint g] []

solve g = solve' gr' where
  aa = areas g
  cp = centrepoint g
  ca = head $ head $ filter (elem cp) aa
  gr' = M.fromList $ map (
    \r1 -> (head r1, map head $ filter (
      \r2 -> head r1 /= head r2 &&
        (not $ null $ intersect (concatMap (neighbours g) r1) r2)
     ) aa
    )
   ) aa
  solve' gr
    | null $ tail $ M.keys $ gr = []
    | otherwise = best : solve' ngr where
      djk _ [] = []
      djk v ((n,q):o) = (n,q) : djk (q:v) (
        o ++ zip (repeat (n+1))
        ((gr M.! q) \\ (v ++ map snd o))
       )
      dout = djk [] [(0,ca)]
      din = let
        m = maximum $ map fst dout
        s = filter ((== m) . fst) dout
        in djk [] s
      rc = filter (flip elem (gr M.! ca) . snd) din
      frc = let
        m = minimum $ map fst rc
        in map snd $ filter ((==m) . fst) rc
      msq = concat $ filter (flip elem frc . head) aa
      clr = map (length &&& head) $ group $ sort $ map (g !) msq
      best = snd $ maximumBy (compare `on` fst) clr
      ngr = let
        ssm = filter ((== best) . (g !)) $ map snd rc
        sml = (concatMap (gr M.!) ssm)
        ncl = ((gr M.! ca) ++ sml) \\ (ca : ssm)
        brk = M.insert ca ncl $ M.filterWithKey (\k _ ->
          (not . flip elem ssm) k
         ) gr
        in M.map 
          (\l -> nub $ map (\e -> if e `elem` ssm then ca else e) l)
          brk


parseFile :: String -> [Grid Word8]
parseFile f = map mk $ filter (not . null . head) $ groupBy ((==) `on` null) $
  map (map ((read :: String -> Word8) . (:[]))) $ lines f where
    mk :: [[Word8]] -> Grid Word8
    mk m = let
      w = fromIntegral (length $ head m) - 1
      h = fromIntegral (length m) - 1
      in array ((0,0),(w,h)) [ ((x,y),v) |
        (y,l) <- zip [h,h-1..] m,
        (x,v) <- zip [0..] l
       ]

showGrid :: Grid Word8 -> String
showGrid g = intercalate "\n" l where
  l = map sl $ groupBy ((==) `on` snd) $
    sortBy ((flip (compare `on` snd)) <> (compare `on` fst)) $
    indices g
  sl = intercalate " " . map (show . (g !))

testsolve = do
  hr <- openFile "floodtest" ReadMode
  sp <- liftM (head . parseFile) $ hGetContents hr
  let
   sol = solve sp
   a = snd $ mapAccumL (\g s -> let g' = applyFill s g in (g',g')) sp sol
  sequence_ $ map (\g -> putStrLn (showGrid g) >> putStrLn "\n") a

C # - 2.383.569

Ini adalah traversal mendalam dari solusi yang mungkin yang secara kasar memilih jalur peningkatan terbaik (mirip / sama dengan entri C Herjan), kecuali saya dengan cerdik membalik urutan kandidat generasi solusi setelah melihat Herjan memposting angka yang sama. Butuh 12+ jam untuk menjalankannya.

class Solver
{
    static void Main()
    {
        int depth = 3;
        string text = File.ReadAllText(@"C:\TEMP\floodtest.txt");
        text = text.Replace("\n\n", ".").Replace("\n", "");
        int count = 0;
        string[] tests = text.Split(new char[] { '.' }, StringSplitOptions.RemoveEmptyEntries);
        for (int i = 0; i < tests.Length; i++)
        {
            Solver s = new Solver(tests[i]);
            string k1 = s.solve(depth);
            count += k1.Length;
            Console.WriteLine(((100 * i) / tests.Length) + " " + i + " " + k1.Length + " " + count + " " + k1);
        }
        Console.WriteLine(count);
    }

    public readonly int MAX_DIM;
    public char[] board;
    public Solver(string prob)
    {
        board = read(prob);
        MAX_DIM = (int)Math.Sqrt(board.Length);
    }

    public string solve(int d)
    {
        var sol = "";
        while (score(eval(copy(board), sol)) != board.Length)
        {
            char[] b = copy(board);
            eval(b, sol);

            var canidates = new List<string>();
            buildCanidates("", canidates, d);
            var best = canidates.Select(c => new {score = score(eval(copy(b), c)), sol = c}).ToList().OrderByDescending(t=>t.score).ThenBy(v => v.sol.Length).First();
            sol = sol + best.sol[0];
        }
        return sol;
    }

    public void buildCanidates(string b, List<string> r, int d)
    {
        if(b.Length>0)
            r.Add(b);
        if (d > 0)
        {
            r.Add(b);
            for (char i = '6'; i >= '1'; i--)
                if(b.Length == 0 || b[b.Length-1] != i)
                    buildCanidates(b + i, r, d - 1);
        }
    }

    public char[] read(string s)
    {
        return s.Where(c => c >= '0' && c <= '9').ToArray();
    }

    public void print(char[] b)
    {
        for (int i = 0; i < MAX_DIM; i++)
        {
            for(int j=0; j<MAX_DIM; j++)
                Console.Write(b[i*MAX_DIM+j]);
            Console.WriteLine();
        }
        Console.WriteLine();
    }

    public char[] copy(char[] b)
    {
        char[] n = new char[b.Length];
        for (int i = 0; i < b.Length; i++)
            n[i] = b[i];
        return n;
    }

    public char[] eval(char[] b, string sol)
    {
        foreach (char c in sol)
            eval(b, c);
        return b;
    }

    public void eval(char[] b, char c)
    {
        foreach (var l in flood(b))
            b[l] = c;
    }

    public int score(char[] b)
    {
        return flood(b).Count;
    }

    public List<int> flood(char[] b)
    {
        int start = (MAX_DIM * (MAX_DIM / 2)) + (MAX_DIM / 2);
        var check = new List<int>(MAX_DIM * MAX_DIM);
        bool[] seen = new bool[b.Length];
        var hits = new List<int>(MAX_DIM*MAX_DIM);

        check.Add(start);
        seen[start]=true;
        char target = b[start];

        int at = 0;
        while (at<check.Count)
        {
            int toCheck = check[at++];
            if (b[toCheck] == target)
            {
                addNeighbors(check, seen, toCheck);
                hits.Add(toCheck);
            }
        }
        return hits;
    }

    public void addNeighbors(List<int> check, bool[] seen, int loc)
    {
        int x = loc / MAX_DIM;
        int y = loc % MAX_DIM;
        addNeighbor(check, seen, x, y - 1);
        addNeighbor(check, seen, x, y + 1);
        addNeighbor(check, seen, x - 1, y);
        addNeighbor(check, seen, x + 1, y);
    }

    public void addNeighbor(List<int> check, bool[] seen, int x, int y)
    {
        if (x >= 0 && x < MAX_DIM && y >= 0 && y < MAX_DIM)
        {
            int l = (x * MAX_DIM) + y;
            if (!seen[l])
            {
                seen[l] = true;
                check.Add(l);
            }
        }
    }
}

Jawa - 2.403.189

BUILD SUCCESSFUL (total time: 220 minutes 15 seconds)

Ini seharusnya menjadi upaya saya pada kekuatan brutal. Tapi! Implementasi pertama saya dari pilihan "terbaik" dalam kedalaman tunggal menghasilkan:

2,589,328 - BUILD SUCCESSFUL (total time: 3 minutes 11 seconds)

Kode yang digunakan untuk keduanya sama dengan brute force yang menyimpan "snapshot" dari gerakan lain yang mungkin dan menjalankan algoritma di atas semuanya.

• Masalah

Jika berjalan dengan pendekatan "multi" pass kegagalan acak akan terjadi. Saya mengatur 100 entri puzzle pertama dalam unit test dan dapat mencapai 100% lulus tetapi tidak 100% dari waktu. Untuk mengimbanginya, saya baru saja melacak nomor puzzle saat ini pada waktu yang gagal dan memulai Thread baru untuk mencari di mana yang terakhir ditinggalkan. Setiap utas menulis hasil masing-masing ke file. Kumpulan file kemudian diringkas menjadi satu file.

• Pendekatan

Nodemewakili ubin / kotak papan dan menyimpan referensi ke semua tetangga itu. Lacak tiga Set<Node>variabel:Remaining , Painted, Targets. Setiap iterasi melihat Targetsuntuk mengelompokkan semua candidatenode berdasarkan nilai, memilih target valuedengan jumlah "terpengaruh" node. Node yang terkena dampak ini kemudian menjadi target untuk iterasi berikutnya.

Sumber tersebar di banyak kelas dan cuplikan tidak terlalu berarti dari konteks keseluruhan. Sumber saya dapat diakses melalui GitHub . Saya juga bermain- main dengan demo JSFiddle untuk visualisasi.

Namun demikian, metode kerja keras saya dari Solver.java:

public void flood() {

 final Data data = new Data();
 consolidateCandidates(data, targets);

 input.add(data.getTarget());

 if(input.size() > SolutionRepository.getInstance().getThreshold()){
  //System.out.println("Exceeded threshold: " + input.toString());
  cancelled = true;
 }
 paintable.addAll(data.targets());
 remaining.removeAll(data.targets());

 if(!data.targets().isEmpty()){
  targets = data.potentialTargets(data.targets(), paintable);

  data.setPaintable(paintable);
  data.setRemaining(remaining);
  data.setInput(input);

  SolutionRepository.getInstance().addSnapshot(data, input);
 }
}

C # - 2.196.462 2.155.834

Ini secara efektif sama dengan pendekatan 'mencari keturunan terbaik' seperti pemecah saya yang lain, tetapi dengan beberapa optimisasi yang nyaris, dengan paralelisme, memungkinkannya mencapai kedalaman 5 dalam waktu kurang dari 10 jam. Dalam proses pengujian ini saya juga menemukan bug di aslinya, sehingga algoritma kadang-kadang akan mengambil rute yang tidak efisien ke keadaan akhir (itu tidak memperhitungkan kedalaman negara dengan skor = 64; ditemukan saat bermain-main dengan hasil kedalaman = 7).

Perbedaan utama antara ini dan pemecah sebelumnya adalah bahwa ia menyimpan Flood Game States dalam memori, sehingga tidak harus membuat ulang 6 ^ 5 status. Berdasarkan penggunaan CPU selama menjalankan, saya cukup yakin ini telah pindah dari CPU terikat ke bandwidth memory terikat. Sangat menyenangkan. Begitu banyak dosa.

Sunting: Karena alasan, algoritme kedalaman 5 tidak selalu menghasilkan hasil terbaik. Untuk meningkatkan kinerja, mari lakukan kedalaman 5 ... dan 4 ... dan 3 dan 2 dan 1, dan lihat mana yang terbaik. Apakah mencukur gerakan 40k lain. Karena kedalaman 5 adalah sebagian besar waktu, menambahkan 4 hingga 1 hanya meningkatkan runtime dari ~ 10 jam hingga ~ 11 jam. Yay!

using System;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Collections.Generic;

public class Program
{
    static void Main()
    {
        int depth = 5;
        string text = File.ReadAllText(@"C:\TEMP\floodtest.txt");
        text = text.Replace("\n\n", ".").Replace("\n", "");
        int count = 0;
        string[] tests = text.Split(new [] { '.' }, StringSplitOptions.RemoveEmptyEntries);

        Stopwatch start = Stopwatch.StartNew();

        const int parChunk = 16*16;
        for (int i = 0; i < tests.Length; i += parChunk)
        {
            //did not know that parallel select didn't respect order
            string[] sols = tests.Skip(i).Take(parChunk).AsParallel().Select((t, idx) => new { s = new Solver2(t).solve(depth), idx}).ToList().OrderBy(v=>v.idx).Select(v=>v.s).ToArray();
            for (int j = 0; j < sols.Length; j++)
            {
                string k1 = sols[j];
                count += k1.Length;
                int k = i + j;
                int estimate = (int)((count*(long)tests.Length)/(k+1));
                Console.WriteLine(k + "\t" + start.Elapsed.TotalMinutes.ToString("F2") + "\t" + count + "\t" + estimate + "\t" + k1.Length + "\t" + k1);
            }
        }
        Console.WriteLine(count);
    }
}

public class Solver2
{
    public readonly int MAX_DIM;
    public char[] board;
    public Solver2(string prob)
    {
        board = read(prob);
        MAX_DIM = (int)Math.Sqrt(board.Length);
    }

    public string solve(int d)
    {
        string best = null;
        for (int k = d; k >= 1; k--)
        {
            string c = subSolve(k);
            if (best == null || c.Length < best.Length)
                best = c;
        }
        return best;
    }

    public string subSolve(int d)
    {
        State current = new State(copy(board), '\0', flood(board));
        var sol = "";

        while (current.score != board.Length)
        {
            State nextState = subSolve(current, d);
            sol = sol + nextState.key;
            current = nextState;
        }
        return sol;
    }

    public State subSolve(State baseState, int d)
    {
        if (d == 0)
            return baseState;
        if (!baseState.childrenGenerated)
        {
            for (int i = 0; i < baseState.children.Length; i++)
            {
                if (('1' + i) == baseState.key) continue; //no point in even eval'ing
                char[] board = copy(baseState.board);
                foreach(int idx in baseState.flood)
                    board[idx] = (char)('1' + i);
                List<int> f = flood(board);
                if (f.Count != baseState.score)
                    baseState.children[i] = new State(board, (char)('1' + i), f);
            }
            baseState.childrenGenerated = true;
        }
        State bestState = null;

        for (int i = 0; i < baseState.children.Length; i++)
            if (baseState.children[i] != null)
            {
                State bestChild = subSolve(baseState.children[i], d - 1);
                baseState.children[i].bestChildScore = bestChild.bestChildScore;
                if (bestState == null || bestState.bestChildScore < bestChild.bestChildScore)
                    bestState = baseState.children[i];
            }
        if (bestState == null || bestState.bestChildScore == baseState.score)
        {
            if (baseState.score == baseState.board.Length)
                baseState.bestChildScore = baseState.score*(d + 1);
            return baseState;
        }
        return bestState;
    }

    public char[] read(string s)
    {
        return s.Where(c => c >= '1' && c <= '6').ToArray();
    }

    public char[] copy(char[] b)
    {
        char[] n = new char[b.Length];
        for (int i = 0; i < b.Length; i++)
            n[i] = b[i];
        return n;
    }

    public List<int> flood(char[] b)
    {
        int start = (MAX_DIM * (MAX_DIM / 2)) + (MAX_DIM / 2);
        var check = new List<int>(MAX_DIM * MAX_DIM);
        bool[] seen = new bool[b.Length];
        var hits = new List<int>(MAX_DIM * MAX_DIM);

        check.Add(start);
        seen[start] = true;
        char target = b[start];

        int at = 0;
        while (at < check.Count)
        {
            int toCheck = check[at++];
            if (b[toCheck] == target)
            {
                addNeighbors(check, seen, toCheck);
                hits.Add(toCheck);
            }
        }
        return hits;
    }

    public void addNeighbors(List<int> check, bool[] seen, int loc)
    {
        //int x = loc / MAX_DIM;
        int y = loc % MAX_DIM;

        if(loc+MAX_DIM < seen.Length)
            addNeighbor(check, seen, loc+MAX_DIM);
        if(loc-MAX_DIM >= 0)
            addNeighbor(check, seen, loc-MAX_DIM);
        if(y<MAX_DIM-1)
            addNeighbor(check, seen, loc+1);
        if (y > 0)
            addNeighbor(check, seen, loc-1);
    }

    public void addNeighbor(List<int> check, bool[] seen, int l)
    {
        if (!seen[l])
        {
            seen[l] = true;
            check.Add(l);
        }
    }
}

public class State
{
    public readonly char[] board;
    public readonly char key;
    public readonly State[] children = new State[6];
    public readonly List<int> flood; 
    public readonly int score;
    public bool childrenGenerated;
    public int bestChildScore;
    public State(char[] board, char k, List<int> flood)
    {
        this.board = board;
        key = k;
        this.flood = flood;
        score = flood.Count;
        bestChildScore = score;
    }
}

Delphi XE3 2.979.145 langkah

Ok jadi ini usaha saya. Saya menyebut bagian pengubah gumpalan, setiap belokan akan membuat salinan array dan menguji setiap warna yang mungkin untuk melihat warna mana yang akan memberikan gumpalan terbesar.

Menjalankan semua teka-teki dalam 3 jam dan 6 menit

program Main;

{$APPTYPE CONSOLE}

{$R *.res}

uses
  SysUtils,
  Classes,
  Generics.Collections,
  math,
  stopwatch in 'stopwatch.pas';

type
  myArr=array[0..1]of integer;
const
  MaxSize=19;
  puzLoc='here is my file';
var
  L:TList<TList<integer>>;
  puzzles:TStringList;
  sc:TList<myArr>;
  a:array[0..MaxSize-1,0..MaxSize-1] of Integer;
  aTest:array[0..MaxSize-1,0..MaxSize-1] of Integer;
  turns,midCol,sX,sY,i:integer;
  currBlob,biggestBlob,ColorBigBlob:integer;
  sTurn:string='';
  GLC:integer=0;

procedure FillArrays;
var
  ln,x,y:integer;
  puzzle:TStringList;
begin
  sc:=TList<myArr>.Create;
  puzzle:=TStringList.Create;    
  while puzzle.Count<19 do
  begin
    if puzzles[GLC]='' then
    begin
      inc(GLC);
      continue
    end
    else
      puzzle.Add(puzzles[GLC]);
    inc(GLC)
  end;    
  for y:=0to MaxSize-1do
    for x:=0to MaxSize-1do
      a[y][x]:=Ord(puzzle[y][x+1])-48;
  puzzle.Free;
end;
function CreateArr(nx,ny:integer):myArr;
begin
  Result[1]:=nx;
  Result[0]:=ny;
end;

procedure CreateBlob;
var
  tst:myArr;
  n,tx,ty:integer;
  currColor:integer;
begin
  n:=0;
  sc.Clear;
  currColor:=a[sy][sx];
  sc.Add(CreateArr(sx,sy));
  repeat
    tx:=sc[n][1];
    ty:=sc[n][0];
    if tx>0 then
      if a[ty][tx-1]=currColor then
      begin
        tst:=CreateArr(tx-1,ty);
        if not sc.Contains(tst)then
          sc.Add(tst);
      end;
    if tx<MaxSize-1 then
      if a[ty][tx+1]=currColor then
      begin
        tst:=CreateArr(tx+1,ty);
        if not sc.Contains(tst)then
          sc.Add(tst);
      end;
    if ty>0 then
      if a[ty-1][tx]=currColor then
      begin
        tst:=CreateArr(tx,ty-1);
        if not sc.Contains(tst)then
          sc.Add(tst);
      end;
    if ty<MaxSize-1 then
      if a[ty+1][tx]=currColor then
      begin
        tst:=CreateArr(tx,ty+1);
        if not sc.Contains(tst)then
          sc.Add(tst);
      end;
    inc(n);
  until (n=sc.Count);
end;

function BlobSize:integer;
var
  L:TList<myArr>;
  tst:myArr;
  n,currColor,tx,ty:integer;
begin
  n:=0;
  L:=TList<myArr>.Create;
  currColor:=aTest[sy][sx];
  L.Add(CreateArr(sx,sy));

  repeat
    tx:=L[n][1];
    ty:=L[n][0];
    if tx>0then
      if aTest[ty][tx-1]=currColor then
      begin
        tst:=CreateArr(tx-1,ty);
        if not L.Contains(tst)then
          L.Add(tst);
      end;
    if tx<MaxSize-1then
      if aTest[ty][tx+1]=currColor then
      begin
        tst:=CreateArr(tx+1,ty);
        if not L.Contains(tst)then
          L.Add(tst);
      end;
    if ty>0then
      if aTest[ty-1][tx]=currColor then
      begin
        tst:=CreateArr(tx,ty-1);
        if not L.Contains(tst)then
          L.Add(tst);
      end;
    if ty<MaxSize-1then
      if aTest[ty+1][tx]=currColor then
      begin
        tst:=CreateArr(tx,ty+1);
        if not L.Contains(tst)then
          L.Add(tst);
      end;
    inc(n);
  until n=l.Count;
  Result:=L.Count;
  L.Free;
end;

function AllsameColor(c:integer):boolean;
var
  cy,cx:integer;
begin
  Result:=true;
  for cy:=0to MaxSize-1do
    for cx:=0to MaxSize-1do
      if a[cy][cx]=c then
        continue
      else
        exit(false);
end;
procedure ChangeColors(old,new:integer; testing:boolean=false);
var
  i,j:integer;
  tst:myArr;
begin
  if testing then
  begin
    for i:= 0to MaxSize-1do
      for j:= 0to MaxSize-1do
        aTest[i][j]:=a[i][j];    
    for I:=0to sc.Count-1do
    begin
      tst:=sc[i];
      aTest[tst[0]][tst[1]]:=new;
    end;
  end
  else
  begin
    for I:=0to sc.Count-1do
    begin
      tst:=sc[i];
      a[tst[0]][tst[1]]:=new;
    end;
  end;
end;
var
  sw, swTot:TStopWatch;
  solved:integer=0;
  solutions:TStringList;
  steps:integer;
  st:TDateTime;
begin          
  st:=Now;
  writeln(FormatDateTime('hh:nn:ss',st));
  solutions:=TStringList.Create;
  puzzles:=TStringList.Create;
  puzzles.LoadFromFile(puzLoc);
  swTot:=TStopWatch.Create(true);
  turns:=0;
  repeat
    sTurn:='';    
    FillArrays;
    sX:=Round(Sqrt(MaxSize))+1;
    sY:=sX;    
    repeat
      biggestBlob:=0;
      ColorBigBlob:=0;
      midCol:=a[sy][sx];
      CreateBlob;
      for I:=1to 6do
      begin
        if I=midCol then continue;    
        ChangeColors(midCol,I,true);
        currBlob:=BlobSize;
        if currBlob>biggestBlob then
        begin
          biggestBlob:=currBlob;
          ColorBigBlob:=i;
        end;
      end;
      ChangeColors(midCol,ColorBigBlob);
      inc(turns);
      if sTurn='' then
        sTurn:=IntToStr(ColorBigBlob)
      else
        sTurn:=sTurn+', '+IntToStr(ColorBigBlob);
    until AllsameColor(a[sy][sx]);
    solutions.Add(sTurn);
    inc(solved);
    if solved mod 100=0then
      writeln(Format('Solved %d puzzles || %s',[solved,FormatDateTime('hh:nn:ss',Now-st)]));    
  until GLC>=puzzles.Count-1;    
  swTot.Stop;
  WriteLn(Format('solving these puzzles took %d',[swTot.Elapsed]));
  writeln(Format('Total moves: %d',[turns]));
  solutions.SaveToFile('save solutions here');
  readln;
end.

Berpikir tentang metode backtracing bruteforce juga.
Mungkin menyenangkan untuk akhir pekan ini ^^

Javascript / node.js - 2.588.847

Algoritme agak berbeda daripada kebanyakan di sini karena menggunakan daerah yang dihitung sebelumnya dan perbedaan status antar perhitungan. Ini berjalan di bawah 10 menit di sini jika Anda khawatir tentang kecepatan karena javascript.

var fs = require('fs')


var file = fs.readFileSync('floodtest','utf8');
var boards = file.split('\n\n');
var linelength  = boards[0].split('\n')[0].length;
var maxdim = linelength* linelength;


var board = function(info){
    this.info =[];
    this.sameNeighbors = [];
    this.differentNeighbors = [];
    this.samedifferentNeighbors = [];
    for (var i = 0;i <info.length;i++ ){
        this.info.push(info[i]|0);
    };

    this.getSameAndDifferentNeighbors();
}

board.prototype.getSameAndDifferentNeighbors = function(){
    var self = this;
    var info = self.info;
    function getSameNeighbors(i,value,sameneighbors,diffneighbors){

        var neighbors = self.getNeighbors(i);
        for(var j =0,nl = neighbors.length; j< nl;j++){
            var index = neighbors[j];
            if (info[index]  === value ){
                if( sameneighbors.indexOf(index) === -1){
                    sameneighbors.push(index);
                    getSameNeighbors(index,value,sameneighbors,diffneighbors);
                }
            }else if( diffneighbors.indexOf(index) === -1){
                    diffneighbors.push(index);
            }
        } 

    }


    var sneighbors = [];
    var dneighbors = [];
    var sdneighbors = [];

    for(var i= 0,l= maxdim;i<l;i++){
        if (sneighbors[i] === undefined){
            var sameneighbors = [i];
            var diffneighbors = [];
            getSameNeighbors(i,info[i],sameneighbors,diffneighbors);
            for (var j = 0; j<sameneighbors.length;j++){
                var k = sameneighbors[j];
                sneighbors[k] = sameneighbors;
                dneighbors[k] = diffneighbors;
            } 
        }

    }

    for(var i= 0,l= maxdim;i<l;i++){
        if (sdneighbors[i] === undefined){
            var value = [];
            var dni = dneighbors[i];
            for (var j = 0,dnil = dni.length; j<dnil;j++){
                var dnij = dni[j];
                var sdnij = sneighbors[dnij];
                for(var k = 0,sdnijl = sdnij.length;k<sdnijl;k++){
                    if (value.indexOf(sdnij[k])=== -1){
                        value.push(sdnij[k]);
                    }
                }
            };
            var sni = sneighbors[i];
            for (var j=0,snil = sni.length;j<snil;j++){
                sdneighbors[sni[j]] = value;
            };
        };
    }
    this.sameNeighbors = sneighbors;
    this.differentNeighbors =  dneighbors;
    this.samedifferentNeighbors =sdneighbors;

}

board.prototype.getNeighbors = function(i){
        var returnValue = [];

        var index = i-linelength;
        if (index >= 0){
            returnValue.push(index);
        }

        index = i+linelength;
        if (index < maxdim){

            returnValue.push(index);
        }

        index = i-1;

        if (index >= 0 && index/linelength >>> 0 === i/linelength  >>> 0){
            returnValue.push(index);
        }
        index = i+1;
        if (index/linelength >>> 0 === i/linelength >>> 0){
            returnValue.push(index);
        }

        if (returnValue.indexOf(-1) !== -1){
            console.log(i,parseInt(index/linelength,10),parseInt(i/linelength,10));
        } 
        return returnValue 
}

board.prototype.solve = function(){
    var i,j;
    var info = this.info;
    var sameNeighbors = this.sameNeighbors;
    var samedifferentNeighbors = this.samedifferentNeighbors;
    var middle = 9*19+9;
    var maxValues = [];

    var done = {};
    for (i=0; i<sameNeighbors[middle].length;i++){
        done[sameNeighbors[middle][i]] = true;
    }
    var usefullNeighbors = [[],[],[],[],[],[],[]];
    var diff = [];
    var count = [0];

    count[1] = 0;
    count[2] = 0;
    count[3] = 0;
    count[4] = 0;
    count[5] = 0;
    count[6] = 0;

    var addusefullNeighbors = function(index,diff){

        var indexsamedifferentNeighbors =samedifferentNeighbors[index];
        for (var i=0;i < indexsamedifferentNeighbors.length;i++){
            var is = indexsamedifferentNeighbors[i];
            var value = info[is];
            if (done[is] === undefined && usefullNeighbors[value].indexOf(is) === -1){
                usefullNeighbors[value].push(is);
                diff.push(value);
            }

        }
    }
    addusefullNeighbors(middle,diff);


    while(  usefullNeighbors[1].length > 0 || usefullNeighbors[2].length > 0 ||
            usefullNeighbors[3].length > 0 || usefullNeighbors[4].length > 0 ||
            usefullNeighbors[5].length > 0 || usefullNeighbors[6].length > 0 ){
        for (i=0;i < diff.length;i++){ 
            count[diff[i]]++;
        };
        var maxValue = count.indexOf(Math.max.apply(null, count));
        diff.length = 0;
        var used = usefullNeighbors[maxValue];
        for (var i=0,ul = used.length;i < ul;i++){
            var index = used[i];
            if (info[index] === maxValue){
                done[index] = true;
                addusefullNeighbors(index,diff);
            }
        }
        used.length = 0;
        count[maxValue] = 0;


        maxValues.push(maxValue);
    }
    return maxValues.join("");
};
var solved = [];
var start = Date.now();
for (var boardindex =0;boardindex < boards.length;boardindex++){ 
    var b = boards[boardindex].replace(/\n/g,'').split('');
    var board2 = new board(b);
    solved.push(board2.solve());
};
var diff = Date.now()-start;
console.log(diff,boards.length);
console.log(solved.join('').length);
console.log("end");

fs.writeFileSync('solution.txt',solved.join('\n'),'utf8');

Kode C yang dijamin untuk menemukan solusi optimal dengan kekuatan kasar sederhana. Bekerja untuk kisi ukuran yang sewenang-wenang dan semua input. Butuh waktu yang sangat, sangat lama untuk dijalankan di sebagian besar grid.

Isi banjir sangat tidak efisien dan bergantung pada rekursi. Mungkin perlu membuat tumpukan Anda lebih besar jika sangat kecil. Sistem brute force menggunakan string untuk menahan angka dan menambahkan-dengan-carry sederhana untuk menggilir semua opsi yang mungkin. Ini juga sangat tidak efisien karena mengulangi sebagian besar langkah segi empat kali.

Sayangnya saya tidak dapat mengujinya terhadap semua kasus uji, karena saya akan mati karena usia tua sebelum selesai.

#include <stdio.h>
#include <string.h>


#define GRID_SIZE       19

char grid[GRID_SIZE][GRID_SIZE] = { {3,3,5,4,1,3,4,1,5,3,3,5,4,1,3,4,1,5},
                                    {5,1,3,4,1,1,5,2,1,3,3,5,4,1,3,4,1,5},
                                    {6,5,2,3,4,3,3,4,3,3,3,5,4,1,3,4,1,5},
                                    {4,4,4,5,5,5,4,1,4,3,3,5,4,1,3,4,1,5},
                                    {6,2,5,3,3,1,1,6,6,3,3,5,4,1,3,4,1,5},
                                    {5,5,1,2,5,2,6,6,3,3,3,5,4,1,3,4,1,5},
                                    {6,1,1,5,3,6,2,3,6,3,3,5,4,1,3,4,1,5},
                                    {1,2,2,4,5,3,5,1,2,3,3,5,4,1,3,4,1,5},
                                    {3,6,6,1,5,1,3,2,4,3,3,5,4,1,3,4,1,5} };
char grid_save[GRID_SIZE][GRID_SIZE];

char test_grids[6][GRID_SIZE][GRID_SIZE];

void flood_fill(char x, char y, char old_colour, char new_colour)
{
    if (grid[y][x] == new_colour)
        return;

    grid[y][x] = new_colour;

    if (y > 0)
    {
        if (grid[y-1][x] == old_colour)
            flood_fill(x, y-1, old_colour, new_colour);
    }
    if (y < GRID_SIZE - 1)
    {
        if (grid[y+1][x] == old_colour)
            flood_fill(x, y+1, old_colour, new_colour);
    }

    if (x > 0)
    {
        if (grid[y][x-1] == old_colour)
            flood_fill(x-1, y, old_colour, new_colour);
    }
    if (x < GRID_SIZE - 1)
    {
        if (grid[y][x+1] == old_colour)
            flood_fill(x+1, y, old_colour, new_colour);
    }
}

bool check_grid(void)
{
    for (char i = 0; i < 6; i++)
    {
        if (!memcmp(grid, &test_grids[i][0][0], sizeof(grid)))
            return(true);
    }

    return(false);
}

void inc_string_num(char *s)
{
    char *c;

    c = s + strlen(s) - 1;
    *c += 1;

    // carry
    while (*c > '6')
    {
        *c = '1';
        if (c == s) // first char
        {
            strcat(s, "1");
            return;
        }
        c--;
        *c += 1;
    }
}

void print_grid(void)
{
    char x, y;
    for (y = 0; y < GRID_SIZE; y++)
    {
        for (x = 0; x < GRID_SIZE; x++)
            printf("%d ", grid[y][x]);
        printf("\n");
    }
    printf("\n");
}

int main(int argc, char* argv[])
{
    // create test grids for comparisons
    for (char i = 0; i < 6; i++)
        memset(&test_grids[i][0][0], i+1, GRID_SIZE*GRID_SIZE);

    char s[256] = "0";
    //char s[256] = "123456123456123455";
    memcpy(grid_save, grid, sizeof(grid));


    print_grid();
    do
    {
        memcpy(grid, grid_save, sizeof(grid));
        inc_string_num(s);

        for (unsigned int i = 0; i < strlen(s); i++)
        {
            flood_fill(4, 4, grid[4][4], s[i] - '0');
        }
    } while(!check_grid());
    print_grid();

    printf("%s\n", s);

    return 0;
}

Sejauh yang saya tahu ini adalah pemenang saat ini. Kompetisi mensyaratkan bahwa:

Program Anda harus sepenuhnya bersifat deterministik; solusi pseudorandom diperbolehkan, tetapi program harus menghasilkan output yang sama untuk test case yang sama setiap waktu.

Memeriksa

Program yang menang akan mengambil jumlah langkah paling sedikit untuk menyelesaikan semua 100.000 kasus uji yang ditemukan dalam file ini (file teks zip, 14,23 MB). Jika dua solusi mengambil jumlah langkah yang sama (misalnya jika mereka berdua menemukan strategi yang optimal), program yang lebih pendek akan menang.

Karena ini selalu menemukan jumlah langkah terendah untuk menyelesaikan setiap papan dan tidak ada yang melakukannya, saat ini ada di depan. Jika seseorang dapat membuat program yang lebih pendek, mereka dapat menang, jadi saya menyajikan versi teroptimalkan ukuran berikut. Eksekusi sedikit lebih lambat, tetapi waktu eksekusi bukan bagian dari kondisi kemenangan:

#include <stdio.h>
#include <string.h>
#define A 9
int g[A][A]={{3,3,5,4,1,3,4,1,5},{5,1,3,4,1,1,5,2,1},{6,5,2,3,4,3,3,4,3},{4,4,4,5,5,5,4,1,4},{6,2,5,3,3,1,1,6,6},{5,5,1,2,5,2,6,6,3},{6,1,1,5,3,6,2,3,6},{1,2,2,4,5,3,5,1,2},{3,6,6,1,5,1,3,2,4}};
int s[A][A];
int t[6][A][A];
void ff(int x,int y,int o,int n)
{if (g[y][x]==n)return;g[y][x]=n;if (y>0){if(g[y-1][x]==o)ff(x,y-1,o,n);}if(y<A-1){if(g[y+1][x]==o)ff(x,y+1,o,n);}if(x>0){if (g[y][x-1] == o)ff(x-1,y,o,n);}if(x<A-1){if(g[y][x+1]==o)ff(x+1,y,o,n);}}
bool check_g(void)
{for(int i=0;i<6;i++){if(!memcmp(g,&t[i][0][0],sizeof(g)))return(true);}return(0);}
void is(char*s){char*c;c=s+strlen(s)-1;*c+=1;while(*c>'6'){*c='1';if (c==s){strcat(s,"1");return;}c--;*c+=1;}}
void pr(void)
{int x, y;for(y=0;y<A;y++){for(x=0;x<A;x++)printf("%d ",g[y][x]);printf("\n");}printf("\n");}
int main(void)
{for(int i=0;i<6;i++)memset(&t[i][0][0],i+1,A*A);char s[256]="0";memcpy(s,g,sizeof(g));pr();do{memcpy(g,s,sizeof(g));is(s);for(int i=0;i<strlen(s);i++){ff(4,4,g[4][4],s[i]-'0');}}while(!check_g());
pr();printf("%s\n",s);return 0;}