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<bits/stdc++.h>
using namespace std;
using LL=long long;
int H,W;
string S[400];  
int c[400][400]={};
vector<pair<LL,LL>>s;
void f(int i,int j,int k=-1){
  if(c[i][j]!=-1){
    assert(k==-1||c[i][j]==k);
    return;
  }
  if(k==-1){
  	c[i][j]=s.size();
  	s.push_back({0,0});
  }else{
    c[i][j]=k;
  }
  
  if(S[i][j]=='#')
    s[c[i][j]].first++;
  else
    s[c[i][j]].second++;
  
  if(i>0  && S[i][j]!=S[i-1][j]) f(i-1,j,c[i][j]);
  if(i<H-1&& S[i][j]!=S[i+1][j]) f(i+1,j,c[i][j]);
  
  if(j>0  && S[i][j]!=S[i][j-1]) f(i,j-1,c[i][j]);
  if(j<W-1&& S[i][j]!=S[i][j+1]) f(i,j+1,c[i][j]);
}
int main(){
  cin>>H>>W;
  for(int i=0;i<H;i++)cin>>S[i];
  
  for(int i=0;i<H;i++)for(int j=0;j<W;j++)c[i][j]=-1;
  for(int i=0;i<H;i++)for(int j=0;j<W;j++)f(i,j);
  
  LL ans=0;
  for(pair<LL,LL> q:s){
    ans+=q.first*q.second;
  }
  cout<<ans<<endl;
}