每日算法 — 使用java实现数墙：约束传播与Flood Fill连通性检测

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游戏简介与算法核心

数墙（Nurikabe）是一款经典的日式逻辑谜题。玩家在矩形网格中涂黑某些格子，满足以下规则：白色格子形成连续的”岛屿”，每个岛屿恰好包含一个数字且格子数等于该数字；黑色格子形成连续的”海洋”，且不能出现2×2的全黑方块。本文用Java实现数墙求解器，讲解约束传播、Flood Fill连通性检测与回溯搜索的结合应用。

数墙规则形式化

将网格视为二维数组，每个格子有三种状态：白（ISLAND）、黑（WATER）、未知（UNKNOWN）。约束条件：

  1. 每个含数字的格子是一个独立岛屿的核心，该岛屿的格子数严格等于数字
  2. 所有白色格子必须属于某个岛屿（岛屿间不相连）
  3. 所有黑色格子必须连通（四连通）
  4. 不存在2×2的黑色格子块
  5. 黑色格子必须形成一个连通的整体

基础数据结构

import java.util.*;

/**
 * 格子状态枚举
 */
enum CellState {
    UNKNOWN('.'),    // 未确定
    ISLAND_WHITE('W'), // 岛屿（白色）
    WATER_BLACK('#');  // 海洋（黑色）

    private final char symbol;
    CellState(char symbol) { this.symbol = symbol; }
    public char getSymbol() { return symbol; }
}

/**
 * 坐标类
 */
class Cell {
    final int row, col;
    Cell(int row, int col) { this.row = row; this.col = col; }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof Cell)) return false;
        Cell c = (Cell) o;
        return row == c.row && col == c.col;
    }

    @Override
    public int hashCode() { return Objects.hash(row, col); }

    @Override
    public String toString() { return "(" + row + "," + col + ")"; }
}

/**
 * 岛屿定义：包含数字的格子及其期望大小
 */
class Island {
    final Cell seed;      // 数字所在格子
    final int targetSize; // 目标大小
    final Set<Cell> cells = new HashSet<>(); // 当前归属该岛屿的格子

    Island(Cell seed, int targetSize) {
        this.seed = seed;
        this.targetSize = targetSize;
        this.cells.add(seed);
    }

    public int currentSize() { return cells.size(); }
    public boolean isComplete() { return currentSize() == targetSize; }
}

/**
 * 数墙棋盘
 */
class NurikabeBoard {
    private final int rows, cols;
    private final CellState[][] grid;
    private final int[][] numbers; // 0表示无数字
    private final List<Island> islands = new ArrayList<>();

    public NurikabeBoard(int[][] puzzle) {
        this.rows = puzzle.length;
        this.cols = puzzle[0].length;
        this.grid = new CellState[rows][cols];
        this.numbers = new int[rows][cols];

        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                grid[r][c] = CellState.UNKNOWN;
                numbers[r][c] = puzzle[r][c];
                if (puzzle[r][c] > 0) {
                    islands.add(new Island(new Cell(r, c), puzzle[r][c]));
                    grid[r][c] = CellState.ISLAND_WHITE; // 数字格必为白
                }
            }
        }
    }

    public CellState get(int r, int c) {
        return grid[r][c];
    }

    public void set(int r, int c, CellState state) {
        grid[r][c] = state;
    }

    public boolean isValidCell(int r, int c) {
        return r >= 0 && r < rows && c >= 0 && c < cols;
    }

    public int getNumber(int r, int c) { return numbers[r][c]; }
    public boolean hasNumber(int r, int c) { return numbers[r][c] > 0; }
    public int getRows() { return rows; }
    public int getCols() { return cols; }
    public List<Island> getIslands() { return islands; }

    /**
     * 获取四邻域格子
     */
    public List<Cell> getNeighbors(int r, int c) {
        List<Cell> neighbors = new ArrayList<>();
        int[][] dirs = {{-1,0}, {1,0}, {0,-1}, {0,1}};
        for (int[] d : dirs) {
            int nr = r + d[0], nc = c + d[1];
            if (isValidCell(nr, nc)) neighbors.add(new Cell(nr, nc));
        }
        return neighbors;
    }

    public NurikabeBoard copy() {
        NurikabeBoard copy = new NurikabeBoard(numbers);
        for (int r = 0; r < rows; r++) {
            System.arraycopy(grid[r], 0, copy.grid[r], 0, cols);
        }
        return copy;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        for (int r = 0; r < rows; r++) {
            for (int c = 0; c < cols; c++) {
                if (numbers[r][c] > 0) sb.append(numbers[r][c]);
                else sb.append(grid[r][c].getSymbol());
                sb.append(' ');
            }
            sb.append('\n');
        }
        return sb.toString();
    }
}

约束传播引擎

/**
 * 约束传播求解器，应用数墙的确定性规则
 */
class ConstraintPropagator {

    /**
     * 执行一轮约束传播，返回是否有新格子被确定
     */
    public boolean propagate(NurikabeBoard board) {
        boolean changed = false;

        // 规则1: 已完成的岛屿，相邻未知格必为黑
        for (Island island : board.getIslands()) {
            if (island.isComplete()) {
                for (Cell cell : island.cells) {
                    for (Cell n : board.getNeighbors(cell.row, cell.col)) {
                        if (board.get(n.row, n.col) == CellState.UNKNOWN) {
                            board.set(n.row, n.col, CellState.WATER_BLACK);
                            changed = true;
                        }
                    }
                }
            }
        }

        // 规则2: 数字格相邻的另一个数字格之间必须涂黑（不同岛屿不能相连）
        for (int r = 0; r < board.getRows(); r++) {
            for (int c = 0; c < board.getCols(); c++) {
                if (board.hasNumber(r, c)) {
                    for (Cell n : board.getNeighbors(r, c)) {
                        if (board.hasNumber(n.row, n.col)) {
                            // 找到两个数字格相邻的中点？不，直接相邻就涂中间？
                            // 实际上数字格本身已经是白色，只需确保它们不被错误连接
                        }
                    }
                }
            }
        }

        // 规则3: 避免出现2×2黑格
        for (int r = 0; r < board.getRows() - 1; r++) {
            for (int c = 0; c < board.getCols() - 1; c++) {
                changed |= preventBlack2x2(board, r, c);
            }
        }

        // 规则4: 检查海洋连通性，隔离的海洋格必须扩展
        changed |= propagateWaterConnectivity(board);

        // 规则5: 岛屿大小约束
        changed |= propagateIslandSize(board);

        return changed;
    }

    /**
     * 防止2×2全黑：如果某2×2区域已有3个黑格，最后一个未知格必为白
     */
    private boolean preventBlack2x2(NurikabeBoard board, int r, int c) {
        boolean changed = false;
        Cell[] square = {
            new Cell(r, c), new Cell(r, c+1),
            new Cell(r+1, c), new Cell(r+1, c+1)
        };

        int blackCount = 0;
        Cell unknown = null;
        for (Cell cell : square) {
            CellState s = board.get(cell.row, cell.col);
            if (s == CellState.WATER_BLACK) blackCount++;
            else if (s == CellState.UNKNOWN) unknown = cell;
        }

        if (blackCount == 3 && unknown != null) {
            board.set(unknown.row, unknown.col, CellState.ISLAND_WHITE);
            changed = true;
        }
        return changed;
    }

    /**
     * 传播海洋连通性约束：确保所有黑格最终能连通
     */
    private boolean propagateWaterConnectivity(NurikabeBoard board) {
        boolean changed = false;
        // 简化版：如果某个未知格被黑格包围且其周围有孤立黑格，可能需涂黑以连接
        // 完整实现需要更复杂的连通分量分析
        return changed;
    }

    /**
     * 传播岛屿大小约束：如果某岛屿还差k格才能达标，
     * 且恰好有k个相邻未知格可达，则这些格必为白
     */
    private boolean propagateIslandSize(NurikabeBoard board) {
        boolean changed = false;

        for (Island island : board.getIslands()) {
            int remaining = island.targetSize - island.currentSize();
            if (remaining <= 0) continue;

            // BFS找可达的未知格
            Set<Cell> reachableUnknown = new HashSet<>();
            Queue<Cell> queue = new LinkedList<>(island.cells);
            Set<Cell> visited = new HashSet<>(island.cells);

            while (!queue.isEmpty()) {
                Cell cur = queue.poll();
                for (Cell n : board.getNeighbors(cur.row, cur.col)) {
                    if (visited.contains(n)) continue;
                    CellState s = board.get(n.row, n.col);
                    if (s == CellState.UNKNOWN) {
                        reachableUnknown.add(n);
                    } else if (s == CellState.ISLAND_WHITE && 
                               !board.hasNumber(n.row, n.col)) {
                        // 属于当前岛屿的白格（非数字格）
                        visited.add(n);
                        queue.add(n);
                    }
                }
            }

            // 如果可达未知格恰好等于剩余需求，全部涂白
            if (reachableUnknown.size() == remaining) {
                for (Cell cell : reachableUnknown) {
                    board.set(cell.row, cell.col, CellState.ISLAND_WHITE);
                    island.cells.add(cell);
                    changed = true;
                }
            }
        }
        return changed;
    }

    /**
     * 执行完整约束传播直到不动点
     */
    public void propagateFully(NurikabeBoard board) {
        while (propagate(board)) {
            // 持续传播直到没有新推断
        }
    }
}

Flood Fill连通性检测

/**
 * Flood Fill连通性检测工具
 */
class ConnectivityChecker {

    /**
     * 检查指定状态的格子是否全部连通（四连通）
     */
    public boolean isConnected(NurikabeBoard board, CellState targetState) {
        // 找到第一个目标状态格子
        Cell start = null;
        outer:
        for (int r = 0; r < board.getRows(); r++) {
            for (int c = 0; c < board.getCols(); c++) {
                if (board.get(r, c) == targetState || 
                    (targetState == CellState.WATER_BLACK && board.get(r, c) == CellState.UNKNOWN)) {
                    start = new Cell(r, c);
                    break outer;
                }
            }
        }
        if (start == null) return true; // 没有该状态格子

        // BFS Flood Fill
        Set<Cell> visited = new HashSet<>();
        Queue<Cell> queue = new LinkedList<>();
        queue.add(start);
        visited.add(start);

        while (!queue.isEmpty()) {
            Cell cur = queue.poll();
            for (Cell n : board.getNeighbors(cur.row, cur.col)) {
                if (visited.contains(n)) continue;
                CellState s = board.get(n.row, n.col);
                // 对黑格连通性检查，未知格也可能是黑
                boolean match = (s == targetState) || 
                    (targetState == CellState.WATER_BLACK && s == CellState.UNKNOWN);
                if (match) {
                    visited.add(n);
                    queue.add(n);
                }
            }
        }

        // 检查是否所有目标状态格子都被访问
        for (int r = 0; r < board.getRows(); r++) {
            for (int c = 0; c < board.getCols(); c++) {
                if (board.get(r, c) == targetState && !visited.contains(new Cell(r, c))) {
                    return false;
                }
            }
        }
        return true;
    }

    /**
     * 检查是否存在2×2全黑区域
     */
    public boolean hasBlack2x2(NurikabeBoard board) {
        for (int r = 0; r < board.getRows() - 1; r++) {
            for (int c = 0; c < board.getCols() - 1; c++) {
                if (isBlack(board, r, c) && isBlack(board, r, c+1) &&
                    isBlack(board, r+1, c) && isBlack(board, r+1, c+1)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean isBlack(NurikabeBoard board, int r, int c) {
        return board.get(r, c) == CellState.WATER_BLACK;
    }

    /**
     * 检查所有岛屿是否满足大小约束且不连通
     */
    public boolean isValidIslands(NurikabeBoard board, List<Island> islands) {
        // 对每个已完成的岛屿，做Flood Fill验证大小
        for (Island island : islands) {
            if (!island.isComplete()) continue;

            Set<Cell> visited = new HashSet<>();
            Queue<Cell> queue = new LinkedList<>();
            queue.add(island.seed);
            visited.add(island.seed);

            while (!queue.isEmpty()) {
                Cell cur = queue.poll();
                for (Cell n : board.getNeighbors(cur.row, cur.col)) {
                    if (visited.contains(n)) continue;
                    CellState s = board.get(n.row, n.col);
                    // 白格且不属于其他数字格
                    if (s == CellState.ISLAND_WHITE) {
                        // 检查是否属于其他岛屿
                        boolean otherIsland = false;
                        for (Island other : islands) {
                            if (other != island && other.seed.equals(n)) {
                                otherIsland = true;
                                break;
                            }
                        }
                        if (!otherIsland) {
                            visited.add(n);
                            queue.add(n);
                        }
                    }
                }
            }

            if (visited.size() != island.targetSize) return false;
        }
        return true;
    }
}

回溯搜索求解器

/**
 * 数墙回溯搜索求解器
 */
class NurikabeSolver {
    private final ConstraintPropagator propagator = new ConstraintPropagator();
    private final ConnectivityChecker checker = new ConnectivityChecker();
    private int steps = 0;

    /**
     * 求解数墙谜题
     */
    public NurikabeBoard solve(NurikabeBoard board) {
        steps = 0;
        // 先进行约束传播简化问题
        propagator.propagateFully(board);

        if (solveRecursive(board)) {
            return board;
        }
        return null;
    }

    private boolean solveRecursive(NurikabeBoard board) {
        steps++;

        // 检查当前状态是否违反约束
        if (checker.hasBlack2x2(board)) return false;
        if (!checker.isConnected(board, CellState.WATER_BLACK)) {
            // 未完成的黑格不连通可能是正常的，需要更精细判断
        }

        // 检查是否全部确定
        if (isComplete(board)) {
            return validateSolution(board);
        }

        // 选择启发式最优的未知格进行分支
        Cell bestCell = selectCell(board);
        if (bestCell == null) return validateSolution(board);

        // 尝试设为白色（岛屿）
        NurikabeBoard copyWhite = board.copy();
        copyWhite.set(bestCell.row, bestCell.col, CellState.ISLAND_WHITE);
        propagator.propagateFully(copyWhite);
        if (solveRecursive(copyWhite)) {
            // 将解复制回原board
            copyGrid(copyWhite, board);
            return true;
        }

        // 尝试设为黑色（海洋）
        NurikabeBoard copyBlack = board.copy();
        copyBlack.set(bestCell.row, bestBlack.col, CellState.WATER_BLACK);
        propagator.propagateFully(copyBlack);
        if (solveRecursive(copyBlack)) {
            copyGrid(copyBlack, board);
            return true;
        }

        return false;
    }

    private Cell selectCell(NurikabeBoard board) {
        // 优先选择约束最强的格子（最少剩余值启发式）
        Cell best = null;
        int minOptions = 3;

        for (int r = 0; r < board.getRows(); r++) {
            for (int c = 0; c < board.getCols(); c++) {
                if (board.get(r, c) == CellState.UNKNOWN) {
                    int options = countValidOptions(board, r, c);
                    if (options < minOptions) {
                        minOptions = options;
                        best = new Cell(r, c);
                    }
                }
            }
        }
        return best;
    }

    private int countValidOptions(NurikabeBoard board, int r, int c) {
        int count = 0;
        // 快速检查设为白/黑是否会立即违规
        if (!wouldViolate2x2(board, r, c, CellState.WATER_BLACK)) count++;
        if (!wouldIsolateIsland(board, r, c)) count++;
        return count;
    }

    private boolean wouldViolate2x2(NurikabeBoard board, int r, int c, CellState state) {
        if (state != CellState.WATER_BLACK) return false;
        // 临时设为黑，检查是否形成2x2
        // 简化检查
        return false;
    }

    private boolean wouldIsolateIsland(NurikabeBoard board, int r, int c) {
        // 如果某数字格的所有相邻未知格被阻断，则无法扩展
        return false;
    }

    private boolean isComplete(NurikabeBoard board) {
        for (int r = 0; r < board.getRows(); r++) {
            for (int c = 0; c < board.getCols(); c++) {
                if (board.get(r, c) == CellState.UNKNOWN) return false;
            }
        }
        return true;
    }

    private boolean validateSolution(NurikabeBoard board) {
        // 检查所有约束
        if (checker.hasBlack2x2(board)) return false;
        if (!checker.isConnected(board, CellState.WATER_BLACK)) return false;

        // 检查每个岛屿大小
        for (Island island : board.getIslands()) {
            Set<Cell> visited = new HashSet<>();
            Queue<Cell> queue = new LinkedList<>();
            queue.add(island.seed);
            visited.add(island.seed);

            while (!queue.isEmpty()) {
                Cell cur = queue.poll();
                for (Cell n : board.getNeighbors(cur.row, cur.col)) {
                    if (visited.contains(n)) continue;
                    if (board.get(n.row, n.col) == CellState.ISLAND_WHITE) {
                        visited.add(n);
                        queue.add(n);
                    }
                }
            }
            if (visited.size() != island.targetSize) return false;
        }

        return true;
    }

    private void copyGrid(NurikabeBoard from, NurikabeBoard to) {
        for (int r = 0; r < from.getRows(); r++) {
            for (int c = 0; c < from.getCols(); c++) {
                to.set(r, c, from.get(r, c));
            }
        }
    }

    public int getSteps() { return steps; }
}

主程序与测试

public class NurikabeMain {
    public static void main(String[] args) {
        // 5×5 简单谜题示例
        // 数字表示岛屿大小，0表示空白
        int[][] puzzle = {
            {5, 0, 0, 0, 0},
            {0, 0, 0, 0, 0},
            {0, 0, 2, 0, 0},
            {0, 0, 0, 0, 0},
            {0, 0, 0, 0, 3}
        };

        System.out.println("=== 数墙谜题 ===");
        NurikabeBoard board = new NurikabeBoard(puzzle);
        System.out.println(board);

        System.out.println("=== 开始求解 ===");
        NurikabeSolver solver = new NurikabeSolver();
        NurikabeBoard solution = solver.solve(board);

        if (solution != null) {
            System.out.println("求解成功！搜索步数: " + solver.getSteps());
            System.out.println(solution);
        } else {
            System.out.println("求解失败");
        }
    }
}

算法复杂度分析

操作 时间复杂度 空间复杂度
约束传播 O(R×C) O(R×C)
Flood Fill O(R×C) O(R×C)
2×2检测 O(R×C) O(1)
回溯搜索（最坏） O(2^(R×C)) O(R×C×深度)
实际带约束传播 远小于最坏情况 O(R×C)

总结

数墙求解器融合了多种经典算法：

  1. 约束传播：利用数墙的确定性规则（如2×2禁止、岛屿大小限制）快速推断，减少搜索空间
  2. Flood Fill：基于BFS的连通性检测，验证海洋连通和岛屿大小约束
  3. 回溯搜索：对不确定的格子进行分支尝试，配合启发式选点（MRV）提高效率
  4. 复制-恢复机制：通过棋盘深拷贝实现搜索状态的高效回溯

约束传播是求解数墙的关键——在典型谜题中，70%以上的格子可以通过纯逻辑推理确定，无需回溯。

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