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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int h = sc.nextInt();
		int w = sc.nextInt();
		char[][] field = new char[h][];
		for (int i = 0; i < h; i++) {
			field[i] = sc.next().toCharArray();
		}
		UnionFindTree uft = new UnionFindTree(h, w);
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				if (j != w - 1 && field[i][j] != field[i][j + 1]) {
					uft.union(j + i * w, j + 1 + i * w, field[i][j], field[i][j + 1]);
				}
				if (i != h - 1 && field[i][j] != field[i + 1][j]) {
					uft.union(j + i * w, j + (i + 1) * w, field[i][j], field[i + 1][j]);
				}
			}
		}
		System.out.println(uft.count());
	}
	
	static class UnionFindTree {
		
		int[] parent;
		int[] bCount;
		int[] wCount;
		int w;
		
		public UnionFindTree(int h, int w) {
			this.w = w;
			parent = new int[h * w];
			bCount = new int[h * w];
			wCount = new int[h * w];
			for (int i = 0; i < h * w; i++) {
				parent[i] = i;
			}
		}
		
		public int find(int i) {
			if (parent[i] == i) {
				return i;
			}
			int p = find(parent[i]);
			parent[i] = p;
			return p;
		}
		
		public int find(int x, int y){
			return find(x + y * w);
		}
		
		public boolean same(int a, int b) {
			return (find(a) == find(b));
		}
		
		public boolean same(int ax, int ay, int bx, int by) {
			return same(ax + ay * w, bx + by * w);
		}
		
		public void union(int a, int b, char aa, char bb) {
			int aaa = find(a);
			int bbb = find(b);
			if (aaa == bbb) {
				return;
			}
			if (bCount[aaa] + wCount[aaa] == 0) {
				if (aa == '#') {
					bCount[aaa] = 1;
				} else {
					wCount[aaa] = 1;
				}
			}
			if (bCount[bbb] + wCount[bbb] == 0) {
				if (bb == '#') {
					bCount[bbb] = 1;
				} else {
					wCount[bbb] = 1;
				}
			}
			parent[aaa] = bbb;
			bCount[bbb] += bCount[aaa];
			wCount[bbb] += wCount[aaa];
		}
		
		public long count() {
			long ans = 0;
			for (int i = 0; i < parent.length; i++) {
				if (parent[i] == i) {
					ans += bCount[i] * wCount[i];
				}
			}
			return ans;
		}
	}
}