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: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
class Program{
	const int mod=1000000007;
	static string[] a;
	static bool[,] h;
	static long ans=0,b=0,c=0;
	static void Main(){
		int[] s=Array.ConvertAll(Console.ReadLine().Split(),int.Parse);
		a=new string[s[0]+2];
		for(int i=1;i<=s[0];i++){a[i]="0"+Console.ReadLine()+"0";}
		a[0]=new string('0',s[1]+2);
		a[s[0]+1]=new string('0',s[1]+2);
		h=new bool[s[0]+2,s[1]+2];
		for(int i=1;i<=s[0];i++){
			for(int j=1;j<=s[1];j++){
				if(!h[i,j]){
					fu(i,j);
					ans+=b*c;
					b=0;c=0;
				}
			}
		}

		Console.WriteLine("{0}",ans);
	}
	static void fu(int y,int x){
		h[y,x]=true;
		if(a[y][x]=='#'){b++;}
		else{c++;}
		if((!h[y-1,x])&&a[y-1][x]!=a[y][x]&&a[y-1][x]!='0'){fu(y-1,x);}
		if((!h[y+1,x])&&a[y+1][x]!=a[y][x]&&a[y+1][x]!='0'){fu(y+1,x);}
		if((!h[y,x-1])&&a[y][x-1]!=a[y][x]&&a[y][x-1]!='0'){fu(y,x-1);}
		if((!h[y,x+1])&&a[y][x+1]!=a[y][x]&&a[y][x+1]!='0'){fu(y,x+1);}
	}
}