HDU2952 Counting Sheep

                                                    Counting Sheep

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1504    Accepted Submission(s): 981

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

 

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解题思路：本题是简单的广度优先搜索题，可视为图的连通性查找，找到大图有几个连通的小图即可。
数羊――>数羊群，遍历整个草原，一旦的发现羊，就把整群都标记了，避免重数。然后，继续遍历，直到整个草原遍历完。

#include<cstdio> #include<cstring> #include<queue> using namespace std; int n,m; char map[100][100]; int sum; int dir[4][2]={1,0,0,1,0,-1,-1,0}; struct node {     int x;     int y; }; int inmap(int x,int y) {     if(x>=0&&x<n&&y>=0&&y<m)         return true;     return false; } void bfs(int x,int y) {     int i;     node u,v;     queue<node> q;     u.x=x;     u.y=y;     q.push(u);     while(!q.empty())     {         u=q.front();         q.pop();         for(i=0;i<4;i++)         {             v.x=u.x+dir[i][0];             v.y=u.y+dir[i][1];             if(inmap(v.x,v.y)&&map[v.x][v.y]=='#')    //这里有一只没有数过的羊             {                 map[v.x][v.y]='*';    //这只羊已经数过了，下次不数它                 q.push(v);             }         }     } } int main() {     int i,j;     int t;     scanf("%d",&t);     while(t--)     {         scanf("%d%d",&n,&m);         sum=0;         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]=='#')    //发现羊                 {                     sum++;                     bfs(i,j);    //找到整群羊，并标记，避免重数                 }             }         }         printf("%d\n",sum);     }     return 0; }