Solve The Maze CodeForces - 1365D（贪心+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).
这个题真的坑爆我了，思路完全正确，但一开始的代码就是wa第8个样例。
思路：对于B来说，我们要在他的四周都设置上‘#’，这样是最贪心的，同样对于结果来说，也是影响最小的。如果B旁边有G的话，那么肯定不能成立。然后我们记录一下G的个数，从(n,m)遍历，如果能遍历的G的个数等于本来的G的个数的话，就说明可以。否则就不可以。
代码如下：

#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;
}

努力加油a啊，(^o)/~