杭电 2120 Ice_cream's world I （并查集求环数）

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 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

Sample Output

3

大意：有n个碉堡，m种关系每个关系表示两个碉堡连在一起，问一共有多少环。
父结点相等就有一个环。

 #include<cstdio>
 int n,m,fa[],i,a,b,sum;
 int find(int a)
 {
     int r=a;
     while(r != fa[r])
     {
         r=fa[r];
     }
     int i=a,j;
     while(i != r)
     {
         j=fa[i];
         fa[i]=r;
         i=j;
     }
     return r;
 }
 void f1(int x,int y)
 {
     int nx,ny;
     nx=find(x);
     ny=find(y);
     if(nx != ny)
     {
         fa[nx]=ny;
     }
     else
     {
         sum++;
     }
 }
 int main()
 {
     while(scanf("%d %d",&n,&m)!=EOF)
     {
         sum=;
         for(i =  ; i < n ; i++)
         {
             fa[i]=i;
         }
         for(i =  ; i < m ; i++)
         {
             scanf("%d %d",&a,&b);
             f1(a,b);
         }
         printf("%d\n",sum);
     }
 }