dfs/bfs--Counting Sheep

Counting Sheep

A while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. Then one day my grandmother suggested I tried counting sheep after I'd gone to bed. As always when my grandmother suggests things, I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed. 

Creative as I am, that wasn't going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps A and C are in the same flock. 


Now, I've got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I've decided I need another computer program that does the counting for me. Then I'll be able to just start both these programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.

Input

The first line of input contains a single number T, the number of test cases to follow. 

Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.

Output

For each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the problem description. 

Notes and Constraints 
0 < T <= 100 
0 < H,W <= 100 

Sample Input

2
4 4
#.#.
.#.#
#.##
.#.#
3 5
###.#
..#..
#.###

Sample Output

6
3

#include<stdio.h>

char field[100][101];
int H,W; //分别代表行数 列数
int T; //代表次数
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1}; //四个方向 


void dfs(int x,int y){
	field[x][y] = '.';
	for(int i = 0;i<4;i++){
		int nx = x+dx[i];
		int ny = y+dy[i];
		if(0<=nx&&nx<H&&0<=ny&&ny<W&&field[nx][ny]=='#')
			dfs(nx,ny);
	}
	return;
} 

int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&H,&W);
		for(int i = 0;i<H;i++){
			scanf("%s",&field[i]);
		}
		int res=0;
		for(int i = 0;i<H;i++)
			for(int j = 0;j<W;j++){
				if(field[i][j]=='#'){
					dfs(i,j);
					res++;
				}
			}
	printf("%d\n",res);
	}
	
}