蒟蒻，大佬勿喷%%

#include<bits/stdc++.h>
using namespace std;

#define int long long
const int N = 1e2+5;
int dx[5] = {0,0,1,-1},
	dy[5] = {1,-1,0,0};
int n,m;
bool vis[N][N],b[N][N],c;

void dfs(int x,int y)
{
	if(x<1 or y<1 or x>n or y>m or b[x][y] or c or vis[x][y]) return;
	if(x == n and y == m)
	{
		c = 1;
		return;
	}
	vis[x][y] = 1;
	for(int i = 0;i<4;i++)
	{
		dfs(x+dx[i],y+dy[i]);
	}
	//vis[x][y] = 0;
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	cin>>n>>m;
	for(int i = 1;i<=n;i++)
		for(int j = 1;j<=m;j++)
		{
			char c;
			cin>>c;
			if(c == '#') b[i][j] = 1;
		}
	dfs(1,1);
	if(c) cout<<"Yes"<<endl;
	else cout<<"No"<<endl;
	return 0;
}

这道题为什么不用回溯qwq，回溯就t了，不回溯就过了