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<vector>
#include<algorithm>
#include<string>

using namespace std;
int dx[4] = {0, 1, 0, -1};
int dy[4] = {1, 0, -1, 0};
bool visited[410][410];
vector<string> s;
int h, w;

bool inside(int a, int b){
    if(a < 0 || a >= h) return false;
    if(b < 0 || b >= w) return false;
    return true;
}

pair<int, int> dfs(int a, int b){
    if(visited[a][b]) return make_pair(0, 0);
    visited[a][b] = true;
    int c1 = 0, c2 = 0;
    if(s[a][b] == '.') c1++;
    else c2++;
    for(int i = 0; i < 4; i++){
        int to_a = a + dy[i], to_b = b + dx[i];
        if(!inside(to_a, to_b)) continue;
        if(s[to_a][to_b] == s[a][b]) continue;
        auto x = dfs(to_a, to_b);
        c1 += x.first;
        c2 += x.second;
    }
    return make_pair(c1, c2);
}

int main(){
    int ans = 0;
    cin >> h >> w;
    s = vector<string>(h);
    for(int i = 0; i < h; i++){
        cin >> s[i];
    }
    for(int i = 0; i < h; i++){
        for(int j = 0; j < w; j++){
            if(s[i][j] != '#') continue;
            auto x = dfs(i, j);
            ans += x.first * x.second;
        }
    }
    cout << ans << endl;
    return 0;
}