用递归求解

/*********************************************************************
    程序名:
    日期: 2026-05-16 16:44
    说明:
*********************************************************************/
#include <bits/stdc++.h>
using namespace std;
char a[103][103];//地图
bool v[103][103];// 标记
int k, n, s1, s2, e1, e2, f;

int fx[] = {-1, 1, 0, 0};

int fy[] = {0, 0, -1, 1};

void dfs(int x, int y) {
	v[x][y] = 1;
	if (————？——) {
		f = ?__;
		return ;
	}
	// x-1,y  x+1,y  x,y-1  x,y+1
//	if(x-1>=1 && x-1<=n && y>=1 &&y<=n && v[x-1][y]==0 && a[x-1][y]!='#' && f==0) dfs(x-1,y);

	for (int i = 0; i <= 3; i++) {
		int tx = x + fx[i];
		int ty = y + fy[i];
		if (__?_)
			dfs(__?_);
	}


	return ;
}

int main() {
	cin >> k;
	for (int t = 1; t <= k; t++) {
		// 每组测试数据

		cin >> n;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				cin >> a[i][j];
				v[i][j] = 0;
			}
		}
		cin >> s1 >> s2 >> e1 >> e2;
		s1++;
		s2++;
		e1++;
		e2++;
		if (a[s1][s2] == '#' || a[e1][e2] == '#') {
			cout << "NO\n";
			continue;
		}
		// s1,s2;
		f = 0;
		dfs(s1, s2);
		if (f == 1) {
			cout << "YES\n";
		} else {
			cout << "NO\n";
		}

	}


	return 0;
}