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 <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll , ll> PLL;
typedef long double ld;

#define pb push_back
#define all(c) c.begin(),c.end()
#define allr(c) c.rbegin(),c.rend()
#define mod 1000000007
#define PI 3.14159265
#define endl '\n'
#define MAX 200001

vector <int> adj[MAX];
int status[MAX];
bool vis[MAX];
int w = 0, b = 0;

void dfs (int s) {
    vis[s] = true;
    w += (!status[s]);
    b += (status[s]);
    for (auto it : adj[s]) {
        if (!vis[it]) {
            dfs(it);
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m;
    cin >> n >> m;
    int value[n][m];
    int arr[n][m];
    int k = 0;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            char c;
            cin >> c;
            value[i][j] = k++;
            arr[i][j] = (c == '#'? 1 : 0);
            status[k - 1] = arr[i][j];
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int v = value[i][j];
            if (arr[i][j]) {
                if (i - 1 >= 0 && !arr[i - 1][j])                    adj[v].pb(value[i - 1][j]) , adj[value[i - 1][j]].pb(v);
                if (i + 1 < n && !arr[i + 1][j])                     adj[v].pb(value[i + 1][j]) , adj[value[i + 1][j]].pb(v);
                if (j - 1 >= 0 && !arr[i][j - 1])                    adj[v].pb(value[i][j - 1]) , adj[value[i][j - 1]].pb(v);
                if (j + 1 < m && !arr[i][j + 1])                     adj[v].pb(value[i][j + 1]) , adj[value[i][j + 1]].pb(v);
            }
        }
    }
    ll ans = 0;
    for (int i = 0; i < n * m; i++) {
        if (!vis[i]) {
            w = 0;
            b = 0;
            dfs(i);
            ans += (b * w);
        }
    }
    cout << ans << endl;

    return 0;
}