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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int n,m,ans,dx[4]={0,0,1,-1},dy[4]={1,-1,0,0},cnt,mp[444][444];
char b[444][444];
bool vis[444][444];
bool ch(int x,int y)
{
	return (x<n && x>=0 && y<m && y>=0);
}
void dfs(int x,int y,int ct,char step)
{
	if(vis[x][y])
	{
		return;
	}
	mp[x][y]=ct;
	vis[x][y]=true;
	if(step=='#')
	{
		step='.';
	}
	else
	{
		step='#';
	}
	for(int i=0;i<4;i++)
	{
		int nx=x+dx[i],ny=y+dy[i];
		if(!ch(nx,ny) || b[nx][ny]!=step)
		{
			continue;
		}
		dfs(nx,ny,ct,step);
	}
}
int main()
{
	cin>>n>>m;
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			cin>>b[i][j];
		}
	}
	for(int i=0;i<n;i++)
	{
		for(int j=0;j<m;j++)
		{
			if(!vis[i][j])
			{
				dfs(i,j,cnt,b[i][j]);
				cnt++;
			}
		}
	}
	for(int k=0;k<cnt;k++)
	{
		int cnt1=0,cnt2=0;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(mp[i][j]==k)
				{
					if(b[i][j]=='.')
					{
						cnt1++;
					}
					else
					{
						cnt2++;
					}
				}
			}
		}
		ans+=cnt1*cnt2;
	}
	cout<<ans<<endl;
	return 0;
}