HDUOJ----2952Counting Sheep

Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1782    Accepted Submission(s): 1170

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

 

Source

IDI Open 2009

简单的搜索...

代码：

//简单的搜索
#include<cstdio>
#include<queue>
#include<iostream>
using namespace std;
const int maxn=101;
char map[maxn][maxn];
typedef struct
{
    int x,y;
}po;
int dir[4][2]={{0,1}, {-1,0}, {1,0} , {0,-1} } ;
int main()
{
    int n,m,t,i,j,k,ans;
    queue<po>tem;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
      scanf("%d%d",&n,&m);
      for(i=0;i<n;i++)
          scanf("%s",map[i]);
      for(i=0;i<n;i++)
      {
          for(j=0;j<m;j++)
          {
              if(map[i][j]=='#')
              {
                  ans++;
                  map[i][j]='.';
                  po st={i,j};
                  tem.push( st );
                  while(!tem.empty())
                  {
                      po en=tem.front();
                      tem.pop();
                      for(k=0;k<4;k++)
                      {
                          if(map[en.x+dir[k][0]][en.y+dir[k][1]]=='#')
                          {
                              map[en.x+dir[k][0]][en.y+dir[k][1]]='.';
                              po sa={en.x+dir[k][0],en.y+dir[k][1]};
                              tem.push(sa);
                          }
                      }
                  }
              }
          }
      }
      printf("%d
",ans);
    }
 return 0;
}