hdu 2952 Counting Sheep

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

 

//简单搜索题

 

#include <iostream>

using namespace std;
char data[105][105];
int n,m;
int to[4][2]={{0,1},{0,-1},{1,0},{-1,0}};

void dfs(int i,int j)
{
  if(i<1||j<1||i>n||j>m) return ;
  for(int k=0;k<4;k++)
  {
      int ii,jj;
      ii=i+to[k][0];
      jj=j+to[k][1];
      if(ii>=1&&ii<=n&&jj>=1&&j<=m&&data[ii][jj]=='#')
      {
          data[ii][jj]='.';
          dfs(ii,jj);
      }
  }
  return ;
}

int main()
{
    int t,ans;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        ans=0;
        for(int i=1;i<=n;i++)
         for(int j=1;j<=m;j++)
         {
             cin>>data[i][j];
         }
        for(int i=1;i<=n;i++)
         for(int j=1;j<=m;j++)
         {
             if(data[i][j]=='#')
             {
                 dfs(i,j);
                 ans++;
             }
         }
        cout<<ans<<endl;
    }
    return 0;
}