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;
#define int long long

struct union_find{
	int N;
	vector<int> parent;
	vector<int> siz;

	union_find(int N) : parent(N), siz(N, 1){
		for (int i = 0; i < N; i++) parent[i] = i;
	}

	int root(int x){
		if(parent[x] == x) return x;
		else return parent[x] = root(parent[x]);
	}

	void merge(int x, int y){
		x = root(x), y = root(y);
		if(x == y) return;
		if(siz[x] < siz[y]) swap(x, y);
		parent[y] = x;
		siz[x] += siz[y];
		siz[y] = 0;
	}

	bool are_same(int x, int y){
		return root(x) == root(y);
	}

	int size(int x){
		return siz[root(x)];
	}
};

signed main(){
	int H, W, i, j;
	scanf("%lld%lld", &H, &W);
	vector<string> S(H);
	for(i = 0; i < H; i++){
		cin >> S[i];
	}
	union_find uf(H * W);
	for(i = 0; i < H; i++){
		for(j = 1; j < W; j++){
			if(S[i][j - 1] != S[i][j]){
				uf.merge(i * W + (j - 1), i * W + j);
			}
		}
	}
	for(i = 1; i < H; i++){
		for(j = 0; j < W; j++){
			if(S[i - 1][j] != S[i][j]){
				uf.merge((i - 1) * W + j, i * W + j);
			}
		}
	}
	vector<int> black(H * W, 0), white(H * W, 0);
	for(i = 0; i < H * W; i++){
		if(S[i / W][i % W] == '#'){
			black[uf.root(i)]++;
		}
		else{
			white[uf.root(i)]++;
		}
	}
	int ans = 0;
	for(i = 0; i < H * W; i++){
		ans += black[i] * white[i];
	}
	printf("%lld\n", ans);
	return 0;
}