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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
class c
{
    static void Main()
    {
        var hw = Array.ConvertAll(Console.ReadLine().Split(), long.Parse);
        var map = new string[hw[0]];
        var check = new long[hw[0]][];
        for (long i = 0; i < hw[0]; i++)
        {
            map[i] = Console.ReadLine();
            check[i] = new long[hw[1]];
        }
        for (long i = 0; i < hw[0]; i++)
        {
            for (int j = 0; j < hw[1]; j++)
            {
                if (check[i][j] == 0)
                {
                    check[i][j] = i * hw[1] + j;
                    if (j > 0 && map[i][j - 1] != map[i][j]) check[i][j] = check[i][j - 1];
                    if (i > 0 && map[i - 1][j] != map[i][j]) check[i][j] = check[i - 1][j];
                    if (i > 0 && j > 0 && (map[i - 1][j] == map[i][j - 1]) && (map[i - 1][j] != map[i][j])) check[i][j - 1] = check[i - 1][j];
                }
            }
        }
        var ansB = new long[hw[0] * hw[1] + 1];
        var ansW = new long[hw[0] * hw[1] + 1];
        for (long i = 0; i < hw[0]; i++)
        {
            for (int j = 0; j < hw[1]; j++)
            {
                if (map[i][j] == '.') ansW[check[i][j]]++;
                else ansB[check[i][j]]++;
            }
        }
        long ans = 0;
        for (long i = 0; i <= hw[0] * hw[1]; i++)
        {
            ans += ansB[i] * ansW[i];
        }
        Console.WriteLine(ans);
    }
}