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 <functional>
#include <queue>
#include <utility>
#include <vector>
#include <iostream>
#include <string>
using namespace std;
typedef pair<int,int> P;

int a[401][401];
int d[401][401];
int h,w;
int dx[4] = {1,0,-1,0};
int dy[4] = {0,1,0,-1};
#define inf 10000000

long long ans,c,nw,sx,sy;

void bfs(int sx,int sy){
    queue<pair<int, int>> q;
    d[sx][sy]=1;
    q.push({sx,sy});
    while(!q.empty()){
        pair<int, int> p = q.front();
        int tx = p.first;
        int ty = p.second;
        q.pop();
        for(int i=0;i<4;i++){
            int x=tx+dx[i];
            int y=ty+dy[i];
            if(x<0 || x>=h)continue;
            if(y<0 || y>=w)continue;
            if((a[tx][ty]!=a[x][y])&&d[x][y]==0){
                c++;
                if(a[tx][ty]==0)nw++;
                d[x][y]=1;
                q.push({x,y});
                
            }
        }
    }
}

int main(void){
    cin>>h>>w;
    string str;
    for(int i=0; i<h; i++){
        cin>>str;
        for(int j=0; j<w; j++){
            if(str[j]=='.')a[i][j]=0;
            else{a[i][j]=1;}
        }
    }
    for(int i=0; i<h; i++){
        for(int j=0; j<w; j++){
            c=1; nw=0;
            if(d[i][j]==0){
                bfs(i,j);
                ans+=nw*(c-nw);
            }
        }
    }
    cout<<ans<<endl;
}