走迷宫

2025.9.15

题目内容

有一个 \(m\times n\) 格的迷宫(表示有 \(m\) 行、\(n\) 列)，其中有可走的也有不可走的，如果用 \(1\) 表示可以走，\(0\) 表示不可以走，读入这 \(m\times n\) 个数据和起始点、结束点（起始点和结束点都是用两个数据来描述的，分别表示这个点的行号和列号）。现在要你编程找出所有可行的路径，要求所走的路径中没有重复的点，走时只能是上下左右四个方向。如果一条路径都不可行，则输出相应信息（用 \(-1\) 表示无路径）

优先顺序：左上右下

输入描述

第一行是两个数 \(1 \le m,n \le 14\)，接下来是 \(m\) 行 \(n\) 列由 \(1\) 和 \(0\) 组成的数据，最后两行是起始点和结束点

输出描述

所有可行的路径，描述一个点时用 \((x,y)\) 的形式，除起始点外，其他的都要用 -> 表示方向

如果没有一条可行的路径则输出 \(-1\)

输入输出样例

输入

5 6
1 0 0 1 0 1
1 1 1 1 1 1
0 0 1 1 1 0
1 1 1 1 1 0
1 1 1 0 1 1
1 1
5 6

输出

(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(2,4)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(3,4)->(4,4)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(2,4)->(2,5)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(3,4)->(3,5)->(4,5)->(5,5)->(5,6)
(1,1)->(2,1)->(2,2)->(2,3)->(3,3)->(4,3)->(4,4)->(4,5)->(5,5)->(5,6)

说明/提示

数据保证随机生成。事实上，如果 \(m=n=14\) 且每个位置都是 \(1\) 的话，有 \(69450664761521361664274701548907358996488\) 种路径

C++实现

#include <bits/stdc++.h>
using namespace std;

int m, n;
vector<vector<int>> maze;
vector<vector<bool>> visited;
vector<pair<int,int>> path;
vector<vector<pair<int,int>>> all_paths;

// 左上右下
int dx[4] = {0, -1, 0, 1};
int dy[4] = {-1, 0, 1, 0};

void dfs(int x, int y, int ex, int ey) {
    if (x == ex && y == ey) {
        all_paths.push_back(path);
        return;
    }

    for (int d = 0; d < 4; d++) {
        int nx = x + dx[d];
        int ny = y + dy[d];
        if (nx >= 1 && nx <= m && ny >= 1 && ny <= n && 
            maze[nx][ny] == 1 && !visited[nx][ny]) {
            visited[nx][ny] = true;
            path.push_back({nx, ny});
            dfs(nx, ny, ex, ey);
            path.pop_back();
            visited[nx][ny] = false;
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> m >> n;
    maze.assign(m+1, vector<int>(n+1));
    visited.assign(m+1, vector<bool>(n+1, false));

    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> maze[i][j];
        }
    }

    int sx, sy, ex, ey;
    cin >> sx >> sy;
    cin >> ex >> ey;

    if (maze[sx][sy] == 0 || maze[ex][ey] == 0) {
        cout << -1 << "\n";
        return 0;
    }

    visited[sx][sy] = true;
    path.push_back({sx, sy});
    dfs(sx, sy, ex, ey);

    if (all_paths.empty()) {
        cout << -1 << "\n";
    } else {
        for (auto &p : all_paths) {
            for (size_t i = 0; i < p.size(); i++) {
                cout << "(" << p[i].first << "," << p[i].second << ")";
                if (i + 1 < p.size()) cout << "->";
            }
            cout << "\n";
        }
    }

    return 0;
}