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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=200010;
const ll mod=1e9+7;

int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

int h,w;
vector<vector<char>> d(410, vector<char>(410,'?'));

int bfs(int x, int y){
  int cnt=0;
  queue<P> q;
  vector<vector<bool>> used(410, vector<bool>(410,false));
  q.push({x,y});
  used[x][y];

  while(!q.empty()){
    P p = q.front(); q.pop();
    rep(i,4){
      int nx = p.F+dx[i], ny = p.S+dy[i];
      if(nx<0 || h<=nx || ny<0 || w<=ny) continue;
      if(used[nx][ny]) continue;
      if(d[p.F][p.S]==d[nx][ny]) continue;
      q.push({nx,ny});
      used[nx][ny] = true;
      if(d[nx][ny]=='.') cnt++;
    }
  }
  return cnt;
}

int main() {
  cin >> h >> w;
  rep(i,h)rep(j,w) cin >> d[i][j];

  ll ans = 0;
  rep(i,h)rep(j,w)if(d[i][j]=='#'){
    ans += bfs(i,j);
  }

  cout << ans << endl;
  return 0;
}