HDU-2120 Ice_cream's world I

 Ice_cream's world I

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

题意：判断成环的个数

#include <iostream>#include <cstdio>const int MAX=1e6+10;int father[MAX];int cout;void init(int n)//初始化 {  for(int i = 0; i <= n; i++)    {     father[i]=i;}} int find(int x)//查找{         if(x!=father[x]) { return father[x]=find(father[x]); }           return x;} int unite(int a,int b) //合并          {          int fa=find(a);          int fb=find(b);          if(fa!=fb)          {          father[fa]=fb;  }else cout++;   return cout; }                     int main(){  int n,m;while(~scanf("%d %d",&n,&m)){          init(n);   int a,b;   cout=0;for(int i = 0;i < m ; i++)  {   scanf(" %d %d",&a,&b);   unite(a,b);  }    printf("%d\n",cout);}return 0;}