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;
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define all(x) (x).begin(),(x).end()
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
typedef long long ll;
const long long INF = 1LL << 60;
typedef pair<int, int> P;

int H, W;
int black;
int white;
vector<string> S;
vector<vector<bool>> used;

void dfs(int x, int y, char n) {
    if (used[y][x] || n == S[y][x]) return;
    if (n == '#' && S[y][x] == '.') white++;
    if (n == '.' && S[y][x] == '#') black++;
    used[y][x] = true;
    if (y + 1 < H) dfs(x, y + 1, S[y][x]);
    if (y - 1 >= 0) dfs(x, y - 1, S[y][x]);
    if (x + 1 < W) dfs(x + 1, y, S[y][x]);
    if (x - 1 >= 0) dfs(x - 1, y, S[y][x]);
}

int main()
{
    cin >> H >> W;
    S = vector<string>(H);
    used = vector<vector<bool>>(H, vector<bool>(W, false));
    rep(i, H) cin >> S[i];

    int ans = 0;
    rep(i, H) {
        rep(j, W) {
            if (used[i][j]) continue;
            dfs(j, i, '-');
            if (S[i][j] == '.') white++;
            else black++;
            ans += black * white;
            white = 0;
            black = 0;
            // cout << i << ":" << j << " " << black << " " << white << endl;
        }
    }
    cout << ans << endl;
    return 0;
}