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 <algorithm>
#include <iostream>
#include <vector>
#include <map>
using namespace std;

struct UnionFind {
    vector<int> data;
    int sz;
    UnionFind(int sz) : data(sz, -1), sz(sz) { }
    bool unionSet(int x, int y) {
        if ((x = root(x)) == (y = root(y))) return false;
        if (data[x] > data[y]) swap(x, y);
        data[x] += data[y]; data[y] = x; sz--;
        return true;
    }
    bool findSet(int x, int y) { return root(x) == root(y); }
    int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
    int size(int x) { return -data[root(x)]; }
    int size() { return sz; }
};

int main() {
    int h, w; cin >> h >> w;
    vector<string> field(h);
    for (string &s: field) cin >> s;
    UnionFind uf(h * w);
    auto idx = [&](int i, int j) { return i * w + j; };
    for (int i = 0; i < h; i++) for (int j = 0; j < w; j++) {
        for (int di: {-1, 0, 1}) for (int dj: {-1, 0, 1}) {
            if (di * di + dj * dj != 1) continue;
            int ni = i + di, nj = j + dj;
            if (ni < 0 || ni >= h || nj < 0 || nj >= w) continue;
            if (field[i][j] != field[ni][nj]) uf.unionSet(idx(i, j), idx(ni, nj));
        }
    }
    vector<vector<int>> cnt(h * w, vector<int>(2, 0));
    for (int i = 0; i < h; i++) for (int j = 0; j < w; j++) {
        cnt[uf.root(idx(i, j))][field[i][j] == '.']++;
    }
    long long ans = 0;
    for (auto &v: cnt) ans += 1LL * v[0] * v[1];
    cout << ans << endl;
    return 0;
}