算法：深搜

题意：让你判断一共有几个羊圈；

思路：像四个方向搜索；

Problem Description

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

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;char a[103][103];int n,m,k;int b[4][2]={0,1,0,-1,-1,0,1,0};void dfs(int x,int y){	a[x][y]='.';	for(int i=0;i<4;i++)	{		int dx=x+b[i][0];		int dy=y+b[i][1];		if(dx>=0&&dx
         
          =0&&dy
          
           >t; while(t--) { k=0; cin>>n>>m; for(i=0;i
           
            >a[i]; for(i=0;i