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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>

int main(){
	int h,w,i,j,count=1,k,flag[400][400]={{0}};
	long ans=0;
	char hw[400][401];
	scanf("%d%d",&h,&w);
	for(i=0;i<h;i++)scanf("%s",hw[i]);
	for(i=0;i<h;i++){
		for(j=0;j<w;j++){
			if(flag[i][j]==0){
				flag[i][j]=count;
				int que[1000][2],st=0,fn=1;
				que[0][0]=i;
				que[0][1]=j;
				while(st<fn){
				if(hw[que[st][0]][que[st][1]]=='#'){
					if(que[st][1]<w-1&&flag[que[st][0]][que[st][1]+1]==0
					&&hw[que[st][0]][que[st][1]+1]=='.'){
						flag[que[st][0]][que[st][1]+1]=count;
						que[fn][0]=que[st][0];
						que[fn++][1]=que[st][1]+1;
					}
					if(que[st][0]<h-1&&flag[que[st][0]+1][que[st][1]]==0
					&&hw[que[st][0]+1][que[st][1]]=='.'){
						flag[que[st][0]+1][que[st][1]]=count;
						que[fn][0]=que[st][0]+1;
						que[fn++][1]=que[st][1];
					}
					if(que[st][1]>0&&flag[que[st][0]][que[st][1]-1]==0
					&&hw[que[st][0]][que[st][1]-1]=='.'){
						flag[que[st][0]][que[st][1]-1]=count;
						que[fn][0]=que[st][0];
						que[fn++][1]=que[st][1]-1;
					}
					if(que[st][0]>0&&flag[que[st][0]-1][que[st][1]]==0
					&&hw[que[st][0]-1][que[st][1]]=='.'){
						flag[que[st][0]-1][que[st][1]]=count;
						que[fn][0]=que[st][0]-1;
						que[fn++][1]=que[st][1];
					}
				}
				if(hw[que[st][0]][que[st][1]]=='.'){
					if(que[st][1]<w-1&&flag[que[st][0]][que[st][1]+1]==0
					&&hw[que[st][0]][que[st][1]+1]=='#'){
						flag[que[st][0]][que[st][1]+1]=count;
						que[fn][0]=que[st][0];
						que[fn++][1]=que[st][1]+1;
					}
					if(que[st][0]<h-1&&flag[que[st][0]+1][que[st][1]]==0
					&&hw[que[st][0]+1][que[st][1]]=='#'){
						flag[que[st][0]+1][que[st][1]]=count;
						que[fn][0]=que[st][0]+1;
						que[fn++][1]=que[st][1];
					}
					if(que[st][1]>0&&flag[que[st][0]][que[st][1]-1]==0
					&&hw[que[st][0]][que[st][1]-1]=='#'){
						flag[que[st][0]][que[st][1]-1]=count;
						que[fn][0]=que[st][0];
						que[fn++][1]=que[st][1]-1;
					}
					if(que[st][0]>0&&flag[que[st][0]-1][que[st][1]]==0
					&&hw[que[st][0]-1][que[st][1]]=='#'){
						flag[que[st][0]-1][que[st][1]]=count;
						que[fn][0]=que[st][0]-1;
						que[fn++][1]=que[st][1];
					}
				}
					st++;
				}
				count++;
			}
		}
	}
	for(k=1;k<count;k++){
		int ck=0,cw=0;
		for(i=0;i<h;i++){
			for(j=0;j<w;j++){
				if(flag[i][j]==k){
					if(hw[i][j]=='#')ck++;
					if(hw[i][j]=='.')cw++;
				}
			}
		}
		ans+=cw*ck;
	}
	printf("%ld\n",ans);
	return 0;
}