反向bfs并记录路径中每个点的前驱，从bfs终点访问每个点的前驱就是答案

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, len;
int g[N][N];
int d[N][N];
pair<int, int> pre[N][N];
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, -0, -1};

void bfs () {
    memset(d, -1, sizeof d);
    queue<pair<int, int>> q;
    d[n-1][n-1] = 0;
    q.push({n-1, n-1});

    while (q.size()) {
        pair<int, int> p = q.front();
        q.pop();
        int px = p.first, py = p.second;

        for (int i = 0; i < 4; i++) {
            int x = px + dx[i], y = py + dy[i];
            if (x >= 0 && y >= 0 && x < n && y < n && g[x][y] == 0 && d[x][y] == -1) {
                d[x][y] = d[px][py] + 1;
                q.push({x, y});
                pre[x][y] = {px, py};
            }
        }
    }
    len = d[0][0];
}

int main(){
    cin >> n;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> g[i][j];
        }
    }
    bfs();
    int x = 0, y = 0;
    cout << x << ' ' << y << '\n';
    for (int i = 0; i < len; i++) {
        pair<int, int> p = pre[x][y];
        x = p.first, y = p.second;
        cout << x << ' ' << y << '\n';

    }
    return 0;
}