Solve The Maze
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.
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 $$$(2, 2)$$$.
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.
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).
import sys
import math
from collections import Counter, defaultdict, deque
from functools import cmp_to_key

def inp():
    return int(input())

def inlt():
    return list(map(int, input().split()))

def insr():
    s = input()
    return list(s[:len(s)])

def invr():
    return map(int, input().split())
t = inp()
for token in range(t):
    (n, m) = invr()
    maze = []
    for blank in range(n):
        maze.append(insr())
    possible = True
    badCount = 0
    goodCount = 0
    for r in range(n):
        for c in range(m):
            if maze[r][c] == 'B':
                badCount += 1
                up = ''
                down = ''
                left = ''
                right = ''
                if r > 0:
                    up = maze[r - 1][c]
                    if up == '.':
                        maze[r - 1][c] = '#'
                if r + 1 < n:
                    down = maze[r + 1][c]
                    if down == '.':
                        maze[r + 1][c] = '#'
                if c > 0:
                    left = maze[r][c - 1]
                    if left == '.':
                        maze[r][c - 1] = '#'
                if c + 1 < m:
                    right = maze[r][c + 1]
                    if right == '.':
                        maze[r][c + 1] = '#'
                if up == 'G' or down == 'G' or left == 'G' or (right == 'G'):
                    possible = False
                    break
            elif maze[r][c] == 'G':
                goodCount += 1
        if not possible:
            break
    if maze[n - 1][m - 1] == '#' and goodCount > 0 or maze[n - 1][m - 1] == 'B':
        possible = False
    if not possible:
        print('No')
        continue
    stack = [(n - 1, m - 1)]
    visited = [[False for _ in range(m)] for _ in range(n)]
    visited[n - 1][m - 1] = True
    goodVisited = 0
    while stack:
        (r, c) = stack.pop()
        if maze[r][c] == 'G':
            goodVisited += 1
        if r > 0 and (not visited[r - 1][c]) and (maze[r - 1][c] == '.' or maze[r - 1][c] == 'G'):
            visited[r - 1][c] = True
            stack.append((r - 1, c))
        if r + 1 < n and (not visited[r + 1][c]) and (maze[r + 1][c] == '.' or maze[r + 1][c] == 'G'):
            visited[r + 1][c] = True
            stack.append((r + 1, c))
        if c > 0 and (not visited[r][c - 1]) and (maze[r][c - 1] == '.' or maze[r][c - 1] == 'G'):
            visited[r][c - 1] = True
            stack.append((r, c - 1))
        if c + 1 < m and (not visited[r][c + 1]) and (maze[r][c + 1] == '.' or maze[r][c + 1] == 'G'):
            visited[r][c + 1] = True
            stack.append((r, c + 1))
    if goodVisited == goodCount:
        print('Yes')
    else:
        print('No')
import sys, random, bisect
from collections import deque, defaultdict
from heapq import heapify, heappop, heappush
from itertools import permutations
from math import gcd, log, sqrt
mi = lambda : map(int, input().split())
li = lambda : list(mi())
for _ in range(int(input())):
    (n, m) = li()
    finish = False
    goods = 0
    A = [[] for _ in range(n)]
    for i in range(n):
        s = input()
        for c in s:
            A[i].append(c)
    for i in range(n):
        for j in range(m):
            if A[i][j] == 'B':
                if i > 0:
                    if A[i - 1][j] == 'G':
                        print('No')
                        finish = True
                        break
                    elif A[i - 1][j] == '.':
                        A[i - 1][j] = '#'
                if i < n - 1:
                    if A[i + 1][j] == 'G':
                        print('No')
                        finish = True
                        break
                    elif A[i + 1][j] == '.':
                        A[i + 1][j] = '#'
                if j > 0:
                    if A[i][j - 1] == 'G':
                        print('No')
                        finish = True
                        break
                    elif A[i][j - 1] == '.':
                        A[i][j - 1] = '#'
                if j < m - 1:
                    if A[i][j + 1] == 'G':
                        print('No')
                        finish = True
                        break
                    elif A[i][j + 1] == '.':
                        A[i][j + 1] = '#'
            elif A[i][j] == 'G':
                goods += 1
        if finish:
            break
    if finish:
        continue
    if A[-1][-1] == 'B':
        print('No')
        continue
    elif goods == 0:
        print('Yes')
        continue
    cua = deque([(n - 1, m - 1)])
    vist = [[False for _ in range(m)] for _ in range(n)]
    while cua:
        (x, y) = cua.popleft()
        if vist[x][y]:
            continue
        vist[x][y] = True
        if A[x][y] == '#':
            continue
        if A[x][y] == 'B':
            print("Hem trobat B? No hauria d'haver passat això")
            break
        if A[x][y] == 'G':
            goods -= 1
        if x > 0 and (not vist[x - 1][y]):
            cua.append((x - 1, y))
        if x < n - 1 and (not vist[x + 1][y]):
            cua.append((x + 1, y))
        if y > 0 and (not vist[x][y - 1]):
            cua.append((x, y - 1))
        if y < m - 1 and (not vist[x][y + 1]):
            cua.append((x, y + 1))
    if goods == 0:
        print('Yes')
    else:
        print('No')
def ss(n, m, l):
    a = []
    b = []
    for i in range(n):
        for j in range(m):
            if l[i][j] == 'B':
                a.append([i, j])
            elif l[i][j] == 'G':
                b.append([i, j])
    if len(b) == 0:
        return 'Yes'
    t = [[0, 1], [1, 0], [-1, 0], [0, -1]]
    for (x, y) in a:
        for (dx, dy) in t:
            nx = x + dx
            ny = y + dy
            if nx < n and ny < m and (nx >= 0) and (ny >= 0):
                if l[nx][ny] == 'G':
                    return 'No'
                if l[nx][ny] == '.':
                    l[nx][ny] = '#'
    if l[n - 1][m - 1] == '#':
        if len(b) == 0:
            return 'Yes'
        return 'No'
    q = [[n - 1, m - 1]]
    while q:
        q1 = []
        for (x, y) in q:
            for (dx, dy) in t:
                nx = x + dx
                ny = y + dy
                if nx < n and ny < m and (nx >= 0) and (ny >= 0) and (l[nx][ny] != '#'):
                    l[nx][ny] = '#'
                    q1.append([nx, ny])
        q = q1
    for i in range(n):
        for j in range(m):
            if l[i][j] == 'G':
                return 'No'
    return 'Yes'
for _ in range(int(input())):
    (n, m) = map(int, input().split())
    l = []
    for i in range(n):
        l.append(list(str(input())))
    print(ss(n, m, l))