Solve The Maze
Contest: Codeforces Round 648 (Div. 2)
Difficulty rating: 1700
Tags: constructive algorithms, dfs and similar, dsu, graphs, greedy, implementation, shortest paths

Vivek has encountered a problem. He has a maze that can be represented as an $$$n \times m$$$ 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 format:
The first line contains one integer $$$t$$$ $$$(1 \le t \le 100)$$$ — the number of test cases. The description of the test cases follows.

The first line of each test case contains two integers $$$n$$$, $$$m$$$ $$$(1 \le n, m \le 50)$$$ — 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 format:
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).

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

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 $$$(2, 2)$$$.

Example input:
6
1 1
.
1 2
G.
2 2
#B
G.
2 3
G.#
B#.
3 3
#B.
#..
GG.
2 2
#B
B.

Example output:
Yes
Yes
No
No
Yes
Yes

Editorial:
Key Idea:
We can block all empty neighbouring cells of bad people and then check if all good people can escape and no bad people are able to escape.
Solution:
Consider all the neighbouring cells of bad people. There shouldn't be any path from these cells to the cell $$$(n, m)$$$. If there is a path from any such cell, the bad person adjacent to that cell can also then reach the cell $$$(n, m)$$$. So, if any good and bad people are in adjacent cells, the answer is "No".
Based on this idea, we can block any empty cell neighbouring a bad person. Suppose there is another solution in which a cell $$$(i, j)$$$ neighbouring a bad person does not need to be blocked. There still won't be any path from $$$(i, j)$$$ to $$$(n, m)$$$ in that solution. So we can block $$$(i, j)$$$ in that solution too without affecting the solution itself.
It is sufficient to block only the empty neighbouring cells of bad people and check the required conditions, which can be done using a bfs on the grid.
Proof:
We will assume there are no adjacent good and bad people since in that case, the answer is "No". There are three cases:
• A bad person is adjacent to the cell $$$(n, m)$$$. In this case, the cell $$$(n, m)$$$ must be blocked. Now no one will be able to escape. If there is at least one good person present, the answer is "No".
• If after blocking the neighbouring cells of bad people, there is some good person who is not able to escape, then the answer is again "No".
• Otherwise, the answer is always "Yes". Suppose there is some path from a bad person at cell $$$(i, j)$$$ to the cell $$$(n, m)$$$. One of the neighbours of this person must be another bad person since the only other case is an adjacent good person (which is already covered above). Extending this, all the cells on the path from $$$(i, j)$$$ to $$$(n, m)$$$ must have bad people. This is not possible since in this case, there must be a bad person adjacent to $$$(n, m)$$$ and this case is already covered above.
Time complexity: $$$O(n \cdot m)$$$