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: Accepted)

#include <bits/stdc++.h>

using namespace std;

int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};

void dfs(int x, int y, long long &cnt_b, long long &cnt_w, const vector<string> &g,
         vector<vector<int>> &vis) {
    int n = g.size(), m = g[0].size();
    vis[x][y] = 1;
    cnt_b += (g[x][y] == '#');
    cnt_w += (g[x][y] == '.');
    for (int i = 0; i < 4; ++i) {
        int nx = x + dx[i], ny = y + dy[i];
        if (nx >= 0 && nx < n && ny >= 0 && ny < m) {
            if (!vis[nx][ny] && g[nx][ny] != g[x][y]) {
                dfs(nx, ny, cnt_b, cnt_w, g, vis);
            }
        }
    }
}

int main() {
    std::ios::sync_with_stdio(false); 
    cin.tie(nullptr);
    
    int H, W;
    cin >> H >> W;
    vector<string> grid(H);
    for (int i = 0; i < H; ++i) {
        cin >> grid[i];
    }
    long long res = 0;
    vector<vector<int>> vis(H, vector<int>(W, 0));
    for (int i = 0; i < H; ++i) {
        for (int j = 0; j < W; ++j) {
            if (!vis[i][j]) {
                long long cur_b = 0, cur_w = 0;
                dfs(i, j, cur_b, cur_w, grid, vis);
                res += cur_b * cur_w;
            }
        }
    }
    cout << res << "\n";

    return 0;
}