Solve The Maze CodeForces - 1365D(욕심+dfs)

Vivek has encountered a problem. He has a maze that can be represented as an n×m grid. Each of the grid cells may represent the following:
Empty — ‘.’ Wall — ‘#’ Good person — ‘G’ Bad person — ‘B’ The only escape from the maze is at cell (n,m).
A person can move to a cell only if it shares a side with their current cell and does not contain a wall. Vivek wants to block some of the empty cells by replacing them with walls in such a way, that all the good people are able to escape, while none of the bad people are able to. A cell that initially contains ‘G’ or ‘B’ cannot be blocked and can be travelled through.
Help him determine if there exists a way to replace some (zero or more) empty cells with walls to satisfy the above conditions.
It is guaranteed that the cell (n,m) is empty. Vivek can also block this cell.
Input The first line contains one integer t (1≤t≤100) — the number of test cases. The description of the test cases follows.
The first line of each test case contains two integers n, m (1≤n,m≤50) — the number of rows and columns in the maze.
Each of the next n lines contain m characters. They describe the layout of the maze. If a character on a line equals ‘.’, the corresponding cell is empty. If it equals ‘#’, the cell has a wall. ‘G’ corresponds to a good person and ‘B’ corresponds to a bad person.
Output For each test case, print “Yes” if there exists a way to replace some empty cells with walls to satisfy the given conditions. Otherwise print “No”
You may print every letter in any case (upper or lower).
Example Input 6 1 1 . 1 2 G. 2 2 #B G. 2 3 G.# B#. 3 3 #B. #… GG. 2 2 #B B. Output Yes Yes No No Yes Yes Note For the first and second test cases, all conditions are already satisfied.
For the third test case, there is only one empty cell (2,2), and if it is replaced with a wall then the good person at (1,2) will not be able to escape.
For the fourth test case, the good person at (1,1) cannot escape.
For the fifth test case, Vivek can block the cells (2,3) and (2,2).
For the last test case, Vivek can block the destination cell (2,2). 이 문제는 정말 나를 곤경에 빠뜨렸다. 생각은 완전히 정확했지만 처음에 코드는wa의 여덟 번째 사례였다.사고방식: B에게 우리는 그의 주위에'#'을 설정해야 한다. 이렇게 하는 것이 가장 욕심이 많고 결과적으로도 영향이 가장 적다.만약 B 옆에 G가 있다면 성립될 수 없을 것이다.그리고 G의 개수를 기록해 봅시다. (n, m)에서 옮겨다니면 G의 개수가 원래의 G의 개수와 같으면 설명할 수 있습니다.안 그러면 안돼.코드는 다음과 같습니다.

#include
#define ll long long
using namespace std;

const int maxx=55;
char s[maxx][maxx];
int vis[maxx][maxx];
int d[][2]={{1,0},{0,1},{-1,0},{0,-1}};
int n,m;

inline int dfs(int x,int y,int &cnt)
{
	vis[x][y]=1;
	for(int i=0;i<4;i++)
	{
		int tx=x+d[i][0];
		int ty=y+d[i][1];
		if(tx<0||tx>=n||ty<0||ty>=m||vis[tx][ty]||s[tx][ty]=='#') continue;
		vis[tx][ty]=1;
		if(s[tx][ty]=='G') cnt++;
		dfs(tx,ty,cnt);
	}
}
inline bool judge()
{
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(s[i][j]=='B')
			{
				for(int t=0;t<4;t++)
				{
					int tx=i+d[t][0];
					int ty=j+d[t][1];
					if(tx<0||tx>=n||ty<0||ty>=m) continue;
					if(s[tx][ty]=='G') return false;
					if(s[tx][ty]=='.'||s[tx][ty]=='#') 	s[tx][ty]='#';
				}
			}
		}
	}
	return true;
}
inline void init()
{
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) vis[i][j]=0;
}
inline bool solve()
{
	int cnt=0,cnt1=0;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++) if(s[i][j]=='G') cnt++;
	}
	if(s[n-1][m-1]=='#'&&cnt) return 0;
	dfs(n-1,m-1,cnt1);
	return cnt1==cnt;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++) scanf("%s",s[i]);
		if(!judge()) 
		{
			cout<<"NO"<<endl;
			continue;
		}
		memset(vis,0,sizeof(vis));
		if(solve()) cout<<"YES"<<endl;
		else cout<<"NO"<<endl;
	}
	return 0;
}

열심히 힘내라 a야,(o)/~