[AcWing 1076] 迷宫问题

BFS 最短路 + 记录方案

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点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1000 + 10;

#define x first
#define y second

int n;
int g[N][N];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
pair<int,int> ne[N][N];

void bfs(int x, int y)
{
    queue<pair<int,int>> q;
    q.push({x, y});
    memset(ne, -1, sizeof ne);
    ne[x][y] = {1, 1};
    while (q.size()) {
        auto t = q.front();
        q.pop();
        for (int i = 0; i < 4; i ++) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 1 || a > n || b < 1 || b > n || g[a][b])
                continue;
            if (ne[a][b].x != -1)
                continue;
            q.push({a, b});
            ne[a][b] = t;
        }
    }
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++)
            cin >> g[i][j];
    bfs(n, n);
    pair<int,int> path(1, 1);
    while (1) {
        cout << path.x - 1 << ' ' << path.y - 1 << endl;
        if (path.x == n && path.y == n)
            break;
        path = ne[path.x][path.y];
    }
}

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

    solve();

    return 0;
}

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1. $BFS$ 时从 $(n, n)$ 点往前搜，对于每个点 $t$，和它能到的点 $p$，让 $ne[p.x][p.y] = t$