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=;
 char map[maxn][maxn];
 typedef struct
 {
     int x,y;
 }po;
 int dir[][]={{,}, {-,}, {,} , {,-} } ;
 int main()
 {
     int n,m,t,i,j,k,ans;
     queue<po>tem;
     scanf("%d",&t);
     while(t--)
     {
         ans=;
       scanf("%d%d",&n,&m);
       for(i=;i<n;i++)
           scanf("%s",map[i]);
       for(i=;i<n;i++)
       {
           for(j=;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=;k<;k++)
                       {
                           if(map[en.x+dir[k][]][en.y+dir[k][]]=='#')
                           {
                               map[en.x+dir[k][]][en.y+dir[k][]]='.';
                               po sa={en.x+dir[k][],en.y+dir[k][]};
                               tem.push(sa);
                           }
                       }
                   }
               }
           }
       }
       printf("%d\n",ans);
     }
  return ;
 }