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<iostream>
#include<string>
using namespace std;
   int w,h;
    char mero[400][400];
    int reached[400][400]={0};
    int a,b;
    long long search(int x,int y){
        if(x<0||h<=x||y<0||w<=y) return 0;
        if(reached[x][y]==1) return 0;
        reached[x][y]=1;
        if(mero[x][y]=='.'){
            a++;
            if(mero[x+1][y]=='#') search(x+1,y);
            if(mero[x-1][y]=='#') search(x-1,y);
            if(mero[x][y+1]=='#') search(x,y+1);
            if(mero[x][y-1]=='#') search(x,y-1);
        }
        if(mero[x][y]=='#'){
            b++;
            if(mero[x+1][y]=='.') search(x+1,y);
            if(mero[x-1][y]=='.') search(x-1,y);
            if(mero[x][y+1]=='.') search(x,y+1);
            if(mero[x][y-1]=='.') search(x,y-1);
        }
    }
int main(){
    cin>>h>>w;
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++) cin>>mero[i][j];
    }
    long long ans=0;
    a=0,b=0;
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
                ans+=a*b;
                a=0,b=0;
                search(i,j);
        }
    }
    ans+=a*b;
    for(int i=0;i<h;i++){
        for(int j=0;j<w;j++){
                ans+=a*b;
                a=0,b=0;
                search(i,j);
        }
    }
    ans+=a*b;
    cout<<ans;
}