还算简单

已AC，放心食用

#include <bits/stdc++.h>
using namespace std;
int n,m,n1,m1,n2,m2,d=0;//n行m列 
char a[100][100];//保存二维数组
int e[4][2]={{0,1},{0,-1},{1,0},{-1,0}};//分别表示上下左右
void fun(int x,int y)
{
	if(d==1)
	{
		return;
	}
	if(x==n2&&y==m2)
	{
		d=1;
		return;
	} 
	for(int i=0;i<4;i++)//循环判断上下左右 
	{
		int x1=x+e[i][0];//计算下一个点的坐标 
		int y1=y+e[i][1];//计算下一个点的坐标 
		if(x1>=0&&y1>=0&&x1<n&&y1<n&&a[x1][y1]!='#')//判断边界，判断是否访问过，判断是否可以通过 
		{
			a[x1][y1]='#';//走过就标记为不能走 
			fun(x1,y1);//进入下一个点
		}
	} 
} 
int main()
{
	cin>>m;
	while(m--)
	{
		cin>>n;
		for(int i=0;i<n;i++)
		{
			for(int k=0;k<n;k++)
			{
				cin>>a[i][k];
			}
		}
		cin>>n1>>m1>>n2>>m2;
		d=0;
		if(a[n1][m1]=='#'||a[n2][m2]=='#')
		{
			cout<<"NO"<<endl;
			continue;
		 } 
		fun(n1,m1);
		if(d==1)
			cout<<"YES"<<endl;
		else
			cout<<"NO"<<endl;
	}
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n,m;
int x,y,x2,y2;
char a[110][110];
int fx[5]={0,0,1,0,-1};
int fy[5]={0,1,0,-1,0};
bool flag;
void dfs(int x,int y){
	a[x][y]='#';
	if (x==x2&&y==y2){
		cout<<"YES"<<endl;
		flag=true;
		return;
	}
	for (int i=1;i<=4;i++){
		int tx=fx[i]+x;
		int ty=fy[i]+y;
		
		if (tx>=0&&tx<m&&ty>=0&&ty<m&&a[tx][ty]=='.'&&!flag)dfs(tx,ty);
	}
}
int main(){
	cin>>n;
	for (int k=1;k<=n;k++){
		flag=false;
		cin>>m;
		for (int i=0;i<m;i++){
			for (int j=0;j<m;j++)cin>>a[i][j];
		}
		cin>>x>>y>>x2>>y2;
		if (a[x][y]=='#'||a[x2][y2]=='#'){
			cout<<"NO"<<endl;
			continue;
		}
		dfs(x,y);
		if (!flag)cout<<"NO"<<endl;
	}
	return 0;
}

#include <iostream>
#include <vector>

bool canReachDestination(std::vector<std::vector<char>>& maze, int n, int ha, int la, int hb, int lb) {
    if (ha < 0 || ha >= n || la < 0 || la >= n || hb < 0 || hb >= n || lb < 0 || lb >= n || maze[ha][la] == '#' || maze[hb][lb] == '#') {
        return false;
    }

    if (ha == hb && la == lb) {
        return true;
    }

    maze[ha][la] = '#';

    if (canReachDestination(maze, n, ha - 1, la, hb, lb) || 
        canReachDestination(maze, n, ha + 1, la, hb, lb) ||
        canReachDestination(maze, n, ha, la - 1, hb, lb) ||
        canReachDestination(maze, n, ha, la + 1, hb, lb)) {
        return true;
    }

    return false;
}

int main() {
    int k;
    std::cin >> k;

    for (int i = 0; i < k; ++i) {
        int n;
        std::cin >> n;

        std::vector<std::vector<char>> maze(n, std::vector<char>(n));
        for (int j = 0; j < n; ++j) {
            for (int l = 0; l < n; ++l) {
                std::cin >> maze[j][l];
            }
        }

        int ha, la, hb, lb;
        std::cin >> ha >> la >> hb >> lb;

        if (canReachDestination(maze, n, ha, la, hb, lb)) {
            std::cout << "YES" << std::endl;
        } else {
            std::cout << "NO" << std::endl;
        }
    }

    return 0;
}

函数AC解法