java - 如何在棋盘中找到最有利可图的路径

Question

假设我们有这样一个板子:

并且我们希望通过以下移动模式从左到右找到最有利可图的路径:

例如，在此版 block 中，最有利可图的路径是:

即{2, 0} -> {2, 1} -> {3, 2} -> {3, 3}

我写了下面的代码:

import java.util.*;

public final class Board {

    private final int[][] board;
    private final int n;

    public Board(int n) {
        board = new int[n][n];
        this.n = n;
        generateBoard();
    }

    public static class Node {

        public int x;
        public int y;
        public int value;

        public Node(int x, int y, int value) {
            this.x = x;
            this.y = y;
            this.value = value;
        }

        @Override
        public String toString() {
            return "{" + x + ", " + y + "}";
        }

    }

    public static class Path implements Comparable<Path> {

        private LinkedList<Node> path = new LinkedList<>();

        public Path() {
        }

        public Path(List<Node> nodes) {
            this.path.addAll(nodes);
        }

        public void addLast(Node node) {
            path.addLast(node);
        }

        public void removeLast() {
            path.removeLast();
        }

        public List<Node> get() {
            return path;
        }

        public int getProfit() {
            return path.stream().map(node -> node.value).mapToInt(value -> value).sum();
        }

        @Override
        public String toString() {
            return path.toString();
        }

        @Override
        public int compareTo(Path o) {
            return getProfit() > o.getProfit() ? 1 : -1;
        }

    }

    public void generateBoard() {
        Random random = new Random();
        for (int x = 0; x < n; x++) {
            for (int y = 0; y < n; y++) {
                board[x][y] = random.nextInt(200) + 1 - 100;
            }
        }
    }

    public void printBoard() {
        for (int[] b : board) {
            System.out.println(Arrays.toString(b));
        }
    }

    public Path findTheMostProfitablePath() {
        TreeSet<Path> paths = new TreeSet<>();
        for (int x = 0; x < n; x++) {
            visit(new Node(x, 0, board[x][0]), paths);
        }
        return paths.last();
    }

    private void visit(Node root, Collection<Path> paths) {
        Stack<Node> stack = new Stack<>();
        stack.add(root);
        Node node;
        Path currentPath = new Path();
        while (!stack.isEmpty()) {
            node = stack.pop();
            currentPath.addLast(node);
            List<Node> children = getChildren(node.x, node.y);
            if (children == null) {
                paths.add(new Path(currentPath.get()));
                currentPath.removeLast();
            } else {
                stack.addAll(children);
            }
        }
    }

    private List<Node> getChildren(int x, int y) {
        if (y == n - 1) {
            return null;
        }
        y++;
        List<Node> children = new LinkedList<>();
        if (x == 0) {
            children.add(new Node(x, y, board[x][y]));
            children.add(new Node(x + 1, y, board[x + 1][y]));
        } else if (x == n - 1) {
            children.add(new Node(x - 1, y, board[x - 1][y]));
            children.add(new Node(x, y, board[x][y]));
        } else {
            children.add(new Node(x - 1, y, board[x - 1][y]));
            children.add(new Node(x, y, board[x][y]));
            children.add(new Node(x + 1, y, board[x + 1][y]));
        }
        return children;
    }

    public static void main(String[] args) {
        Board board = new Board(3);
        System.out.println("Board :");
        board.printBoard();
        System.out.println("\nThe most profitable path :\n" + board.findTheMostProfitablePath());
    }

}

但是找不到路径... 输出:

Board :
[-7, 1, 18]
[88, 56, 18]
[-18, -13, 100]

The most profitable path :
[{1, 0}, {2, 1}, {1, 1}, {2, 2}]

我的代码有什么问题？

我知道这不是找到最赚钱路径的最佳算法，而且速度很慢。 在 n*n 板中，路径的数量为:

n * 2 ^ (n-1) < number of paths < n * 3 ^ (n-1)

在这个算法中，我们正在检查所有路径以找到最有利可图的路径。

谢谢。

Answer 1

您正在解决 Longest Path Problem ，通常是 NP-Complete。但是，在您的情况下，您实际上有一个 Directed Acyclic Graph ，因此可以使用以下递推公式获得有效的解决方案:

D(i,-1) = D(i,m) = -INFINITY //out of bounds
D(-1,j) = 0   [  j>=0  ]     //succesful path
D(i,j) = max{ D(i-1, j-1) , D(i-1,j),D(i-1,j+1)} + A[i][j] //choose best out of 3

申请Dynamic Programming实现此公式的技术，可以实现高效的 O(n*m) 最优解。

当你完成后，在最右边的列上进行简单扫描就会找到“最有利可图”路径的目的地，你可以通过返回上面的内容并写下每个决策来从那里重建实际路径步骤。