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.*;
import java.awt.*;
import java.awt.geom.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
    public static void main(String[] $) {
        Scanner sc = new Scanner(in);
        int h=sc.nextInt();
        int w=sc.nextInt();
        char[][] f=new char[h][w];
        for (int i = 0; i < h; i++) {
            String t=sc.next();
            for (int j = 0; j < w; j++) {
                f[i][j]=t.charAt(j);
            }
        }
        int ans=0;
        int[] dx={-1,0,1,0};
        int[] dy={0,-1,0,1};
        int[][] node=new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                int tempW=0,tempB=0;
                if(node[i][j]!=0)break;
                Queue<Point> q=new ArrayDeque<>();
                q.add(new Point(i,j));
                while (!q.isEmpty()) {
                    Point p = q.poll();
                    if(f[p.x][p.y]=='.')tempW++;
                    else tempB++;
                    node[p.x][p.y] = 1;
                    for (int k = 0; k < 4; k++) {
                        int x = p.x + dx[k];
                        int y = p.y + dy[k];
                        if (0 <= x && x < h && 0 <= y && y < w) {
                            if (node[x][y] == 0 && f[p.x][p.y] != f[x][y]&&!q.contains(f[p.x][p.y])) {
                                q.add(new Point(x, y));
                            }
                        }
                    }
                }
                ans+=tempB*tempW;
            }
        }
        out.println(ans);
    }
}