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: Time Limit Exceeded)

#include<stdio.h>
int count;
long long int cb, cw;
int h, w;
char s[402][402];
int visited[402][402];
void dfs(int x, int y, int k)
{
	if (visited[x][y] > 0)
		return;
	visited[x][y] = k;
	count++;
	if (s[x][y] == '#')
		cb++;
	else
		cw++;
	if (x > 0)
		if (s[x][y] != s[x - 1][y])
			dfs(x - 1, y, k);
	if (y > 0)
		if (s[x][y] != s[x][y - 1])
			dfs(x, y - 1, k);
	if (x < h - 1)
		if (s[x][y] != s[x + 1][y])
			dfs(x + 1, y, k);
	if (y < w - 1)
		if (s[x][y] != s[x][y + 1])
			dfs(x, y + 1, k);
	return;
}
int main()
{
	scanf("%d %d", &h, &w);
	int i, j;
	for (i = 0; i < h; i++)
		scanf("%s", s[i]);
	for (i = 0; i < h; i++)
		for (j = 0; j < w; j++)
			visited[i][j] = 0;
	int k = 1;
	int x, y;
	long long int ans = 0;
	while (count < h * w)
	{
		cw = cb = 0;
		x = y = 0;
		for (i = 0; i < h; i++)
		{
			for (j = 0; j < w; j++)
			{
				if (visited[i][j] == 0)
				{
					x = i;
					y = j;
					break;
				}
			}
			if (visited[x][y] == 0)
				break;
		}
		dfs(x, y, k);
		ans += cw * cb;
		k++;
	}
	printf("%lld\n", ans);
	return 0;
}