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[1005],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=0;
        for(i = 0 ; i < n ; i++)
        {
            fa[i]=i;
        }
        for(i = 0 ; i < m ; i++)
        {
            scanf("%d %d",&a,&b);
            f1(a,b);
        }
        printf("%d
",sum);
    }
}