Ice_cream's world I<hdoj2120>

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

 

Sample Input

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

       很简单的题     查找环的个数

#include<stdio.h>#include<cstring>int pre[1001];int find(int p)  //路径压缩 {      int r=p;      int t;      while(p!=pre[p])      p=pre[p];      while(r!=p)      {          t=pre[r];          pre[r]=p;          r=t;      }      return p;  } void merge(int x,int y){int fx=find(x);int fy=find(y);if(fx!=fy){pre[fx]=fy;}}int main(){int m,n,a,b,i,flog,k,f;while(scanf("%d%d",&a,&b)!=EOF){for(int i=0;i<a;i++)    {pre[i]=i;     }        flog=0;       while(b--)   {   scanf("%d%d",&n,&m);         if(find(n)==find(m))//判断是否成环          flog++;         merge(n,m);}    printf("%d\n",flog);}return 0; }