Problem:
Score : 300 points

Problem Statement

There is a grid with H rows and W columns, where each square is painted black or white.

You are given H strings S_1, S_2, ..., S_H, each of length W.
If the square at the i-th row from the top and the j-th column from the left is painted black, the j-th character in the string S_i is #; if that square is painted white, the j-th character in the string S_i is ..

Find the number of pairs of a black square c_1 and a white square c_2 that satisfy the following condition:

There is a path from the square c_1 to the square c_2 where we repeatedly move to a vertically or horizontally adjacent square through an alternating sequence of black and white squares: black, white, black, white...

Constraints

1 \leq H, W \leq 400

|S_i| = W (1 \leq i \leq H)

For each i (1 \leq i \leq H), the string S_i consists of characters # and ..

Input

Input is given from Standard Input in the following format:

H W
S_1
S_2
:
S_H

Output

Print the answer.

Sample Input 1

3 3
.#.
..#
#..

Sample Output 1

10

Some of the pairs satisfying the condition are ((1, 2), (3, 3)) and ((3, 1), (3, 2)), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

Sample Input 2

2 4
....
....

Sample Output 2

0

Sample Input 3

4 3
###
###
...
###

Sample Output 3

6

Solution in PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys
sys.setrecursionlimit(10**7)
black = 0
white = 0
#x,y渡すと、（ｈ、ｗ）に入ってるかどうか判定
def check(s,t):
    if (0 <=  s <= h - 1) and (0 <= t <= w - 1):
        return True
    else:
        return False
def bfs(que):
    finished = set()
    while que:
        x , y, color = que.popleft()
        if x+y*w not in finished:
            finished.add(x+y*w)
            for i, j in ((1, 0), (0, 1)):
                if abs(i) + abs(j) < 2 and check(y+j,x+i):
                    if (x+i)+(y+j)*w not in finished:
                        que.append([x+i,y+j,grid[y+j][x+i]])
                    if color != grid[y+j][x+i]:
                        adjacent_list[x+w*y].append((x + i)+w*(y + j))
def dfs(node):
    global white
    global black
    finished.add(node)
    y,x = divmod(node,w)
    if grid[y][x] == '#':
        black += 1
    else:
        white += 1
    for i in adjacent_list[node]:
        if i not in finished:
            dfs(i)
from collections import deque
h,w = map(int, input().split())
grid = []
for i in range(h):
    grid.append(list(input()))
ans = 0
adjacent_list = [[] for i in range(h*w)]
que = deque()
que.append([0,0,grid[0][0]])
bfs(que)
#再び初期化
finished = set()
for i in range(h*w):
    #頂点iからdfsをする。黒と白をカウントしておいて最後にかける
    black = 0
    white = 0
    if i not in finished:
        dfs(i)
        ans += black * white
print(ans)