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.

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.

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

 1 #include<iostream>
 2 using namespace std;
 3 
 4 int arr[4][2]={0,1,1,0,0,-1,-1,0};    //分四个方向搜索 
 5 char sheep[101][101];
 6 int T,i,j,sum,h,w;
 7 
 8 //搜索羊群 
 9 void bfs(int a,int b)
10 {
11     if(sheep[a][b]==‘.‘)    return;            //遇到 ‘.‘返回 
12     if(a<0||b<0||a>=h||b>=w)    return;        //数组越界返回 
13     sheep[a][b]=‘.‘;                        //记录，将每次找到的羊群变成‘.‘，下次循环直接跳过 
14     for(int i=0;i<4;i++)
15         bfs(a+arr[i][0],b+arr[i][1]);        //找到每只羊以后向四个方向搜索 
16 }
17 
18 int main()
19 {    
20     cin>>T;
21     while(T--)
22     {
23         cin>>h>>w;
24         for(i=0;i<h;i++)
25             for(j=0;j<w;j++)
26                  cin>>sheep[i][j];        //此处利用c++的输入方法，不用考虑缓冲区换行符的影响 
27         sum=0;
28         for(i=0;i<h;i++)
29             for(j=0;j<w;j++)
30             {
31                 if(sheep[i][j]==‘#‘)
32                 {
33                     ++sum;                //用sum来记录每次找到的羊群数量 
34                     bfs(i,j);
35                 }
36             }
37         cout<<sum<<endl;
38     }
39     return 0;
40 }