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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <utility>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
const ll MOD = 1e9+7;
string s[400];
bool used[400][400];

P operator+(const P& lhs,const P& rhs)
{
	return P(lhs.first+rhs.first,lhs.second+rhs.second);
}

P dfs(P now,int h,int w)
{
	//cerr<<now.first<<" "<<now.second<<endl;
	P ret=P(0,0);
	used[now.first][now.second]=true;
	bool f;
	if(s[now.first][now.second]=='.')
	{
		ret.second++;
		f=false;
	}else{
		ret.first++;
		f=true;
	}
	int dx[4]={1,0,-1,0};
	int dy[4]={0,-1,0,1};
	for(int i=0;i<4;i++)
	{
		if(now.first+dx[i]>-1 && now.first+dx[i]<h && now.second+dy[i]>-1 && now.second+dy[i]<w)
		{
			if(!used[now.first+dx[i]][now.second+dy[i]])
			{
				char c=s[now.first+dx[i]][now.second+dy[i]];
				if((c=='.' && f) || (c=='#' && !f))
					ret=ret+dfs(P(now.first+dx[i],now.second+dy[i]),h,w);
			}
		}
	}
	return ret;
}

int main()
{
	int h,w;
	ll ans=0;
	cin>>h>>w;
	for(int i=0;i<h;i++)
		cin>>s[i];
	fill(used[0],used[400],false);
	for(int i=0;i<h;i++)
	{
		for(int j=0;j<w;j++)
		{
			if(!used[i][j])
			{
				P p=dfs(P(i,j),h,w);
				ans+=p.first*p.second;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}