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).
這個題真的坑爆我了，思路完全正確，但一開始的代碼就是wa第8個樣例。
思路：對于B來說，我們要在他的四周都設置上‘#’，這樣是最貪心的，同樣對于結果來說，也是影響最小的。如果B旁邊有G的話，那么肯定不能成立。然后我們記錄一下G的個數，從(n,m)遍歷，如果能遍歷的G的個數等于本來的G的個數的話，就說明可以。否則就不可以。
代碼如下：

#include<bits/stdc++.h> #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)/~

總結

以上是生活随笔為你收集整理的Solve The Maze CodeForces - 1365D（贪心+dfs）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。