#include<bits/stdc++.h>
using namespace std;
int dx[5] {0, 1, -1, 0, 0};
int dy[5] {0, 0, 0, 1, -1};
int mapp[1005][1005];
int n, desx, desy;
const int inf = 2147483647;
bool vis[1005][1005];
char der[5] {'\0', 'd', 'u', 'r', 'l'};
int dfs(int x, int y) {
	int ret = inf;
	if (mapp[x][y]) {
	//	printf("(%d,%d)是死路，不能走\n",x,y);
		return inf;   //走回头路或者到店铺就惩罚
	}
	if(vis[x][y])return inf; 
	vis[x][y]=1;
	if (desx == x && desy == y)return 0; //到终点就返回0
	for (int i = 1; i <= 4; i++) {
		int nx = x + dx[i];
		int ny = y + dy[i];
		if (nx <= 0 || nx > n || ny <= 0 || ny > n) {
			continue; //越界就下一个
		}
		int now=dfs(nx,ny);
		ret = min(now, ret); //求四个的最小值
	//	if(now!=inf)printf("从(%d,%d)向%c走到(%d,%d)\n", x, y, der[i], nx, ny);
	}
	if (ret == inf)return inf; //自己是思路&四个方向全是死路就惩罚
	return ret + 1; //返回下一步的最小值+自己

}
int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		string s;
		cin >> s;
		for (int j = 1; j <= n; j++)
			mapp[i][j] = s[j - 1] - '0';
	}
	int sx, sy;
	cin >> sx >> sy >> desx >> desy;
	cout << dfs(sx, sy);
}

或者谁能给一个我这个程序过不了的数据（这个洛谷他不到点不给我获取次数）

谢谢各位大佬