题解 | #走迷宫#

#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <tuple>

using namespace std;

int main() {
    int n, m;
    cin >> n >> m;

    int sx, sy, ex, ey;
    cin >> sx >> sy >> ex >> ey;
    // 转换为 0-based 索引
    sx--; sy--; ex--; ey--;

    vector<string> grid(n);
    for (int i = 0; i < n; ++i) {
        cin >> grid[i];
    }

    vector<vector<int>> dist(n, vector<int>(m, -1));
    queue<tuple<int, int, int>> q;

    // 起点初始化
    dist[sx][sy] = 0;
    q.push({sx, sy, 0});

    // 定义四个方向
    int dx[] = {-1, 1, 0, 0};
    int dy[] = {0, 0, -1, 1};

    while (!q.empty()) {
        int x, y, d;
        tie(x, y, d) = q.front();
        q.pop();

        if (x == ex && y == ey) {
            cout << d << endl;
            return 0;
        }

        for (int i = 0; i < 4; ++i) {
            int nx = x + dx[i];
            int ny = y + dy[i];

            // 检查新坐标的合法性
            if (nx >= 0 && nx < n && ny >= 0 && ny < m &&
                grid[nx][ny] == '.' && dist[nx][ny] == -1) {
                
                dist[nx][ny] = d + 1;
                q.push({nx, ny, d + 1});
            }
        }
    }

    cout << -1 << endl;

    return 0;
}

import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        int sx = sc.nextInt() - 1;
        int sy = sc.nextInt() - 1;
        int ex = sc.nextInt() - 1;
        int ey = sc.nextInt() - 1;

        char[][] grid = new char[n][m];
        for (int i = 0; i < n; i++) {
            grid[i] = sc.next().toCharArray();
        }

        int[][] dist = new int[n][m];
        for (int[] row : dist) {
            Arrays.fill(row, -1);
        }

        Queue<int[]> q = new LinkedList<>();

        dist[sx][sy] = 0;
        q.offer(new int[]{sx, sy, 0});

        int[] dx = {-1, 1, 0, 0};
        int[] dy = {0, 0, -1, 1};

        while (!q.isEmpty()) {
            int[] current = q.poll();
            int x = current[0];
            int y = current[1];
            int d = current[2];

            if (x == ex && y == ey) {
                System.out.println(d);
                return;
            }

            for (int i = 0; i < 4; i++) {
                int nx = x + dx[i];
                int ny = y + dy[i];

                if (nx >= 0 && nx < n && ny >= 0 && ny < m &&
                    grid[nx][ny] == '.' && dist[nx][ny] == -1) {
                    
                    dist[nx][ny] = d + 1;
                    q.offer(new int[]{nx, ny, d + 1});
                }
            }
        }

        System.out.println(-1);
    }
}

import sys
from collections import deque

def main():
    # 读取输入
    n, m = map(int, sys.stdin.readline().split())
    sx, sy, ex, ey = map(int, sys.stdin.readline().split())
    # 转换为 0-based 索引
    sx, sy, ex, ey = sx - 1, sy - 1, ex - 1, ey - 1

    grid = [sys.stdin.readline().strip() for _ in range(n)]

    # 初始化距离/访问数组
    dist = [[-1] * m for _ in range(n)]
    
    # 初始化队列
    q = deque()
    
    dist[sx][sy] = 0
    q.append((sx, sy, 0))

    # 四个方向
    dx = [-1, 1, 0, 0]
    dy = [0, 0, -1, 1]

    while q:
        x, y, d = q.popleft()

        if x == ex and y == ey:
            print(d)
            return

        for i in range(4):
            nx, ny = x + dx[i], y + dy[i]

            if 0 <= nx < n and 0 <= ny < m and \
               grid[nx][ny] == '.' and dist[nx][ny] == -1:
                
                dist[nx][ny] = d + 1
                q.append((nx, ny, d + 1))

    print(-1)

if __name__ == "__main__":
    main()