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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {

  long h,w;
  scanf("%ld %ld", &h, &w);
  char s[h][w+1];
  for (long i = 0; i < h; i++) {
    scanf("%s", s[i]);
  }
  long visited[h*w],queue[h*w];
  for (long i = 0; i < h*w; i++) {
    visited[i] = 0;
    queue[i] = -1;
  }
  visited[0] = 1;
  queue[0] = 0;
  long head = 0, tail = 0, start = 0, end = 0, cur, flag;
  long ans = 0;
  long black = 0, white = 0, all = 1;
  if (s[0][0] == '#') {
    black++;
  } else {
    white++;
  }
  while (1) {
    flag = 0;
    for (long i = start; i <= end; i++) {
      cur = queue[i];
      if (cur%w != w-1 && visited[cur+1] == 0 && s[cur/w][cur%w] != s[cur/w][cur%w+1]) {
        if (s[cur/w][cur%w+1] == '#') {
          black++;
        } else {
          white++;
        }
        all++;
        visited[cur+1] = 1;
        tail++;
        queue[tail] = cur+1;
        flag = 1;
      }
      if (cur/w != h-1 && visited[cur+w] == 0 && s[cur/w][cur%w] != s[cur/w+1][cur%w]) {
        if (s[cur/w+1][cur%w] == '#') {
          black++;
        } else {
          white++;
        }
        all++;
        visited[cur+w] = 1;
        tail++;
        queue[tail] = cur+w;
        flag = 1;
      }
    }
    if (flag == 0) {
      ans += black*white;
      if (tail == h*w-1) {
        break;
      } else {
        black = 0;
        white = 0;
        for (long i = head+1; i < h*w; i++) {
          if (visited[i] == 0) {
            visited[i] = 1;
            head = i;
            break;
          }
        }
        tail++;
        queue[tail] = head;
        if (s[head/w][head%w] == '#') {
          black++;
        } else {
          white++;
        }
        all++;
        start = tail;
        end = tail;
      }
    } else {
      start = end+1;
      end = tail;
    }
  }
  if (all != h*w) {
    return -1;
  }
  printf("%ld\n", ans);

  return 0;
}