poj 1611 求0号结点所在集合的元素个数

求0号结点所在集合的元素个数

Sample Input

100 4
2 1 2
5 10 13 11 12 14
2 0 1
2 99 2
200 2
1 5
5 1 2 3 4 5
1 0
0 0
Sample Output

4
1
1

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <cmath>
# include <queue>
# define LL long long
using namespace std ;

const int MAXN = 30010;
int F[MAXN];
int num[MAXN] ;

int find(int x)//找x的祖先结点
{
    if(F[x]==x) return x;
    return F[x]=find(F[x]);
}
void bing(int u,int v) //按秩合并
{
    int x = find(u);
    int y = find(v);
    if(x == y)
        return ;
    if(num[x] >= num[y])
    {
        F[y] = x;
        num[x] += num[y];
    }
    else
    {
        F[x] = y;
        num[y] += num[x];
    }
}
int main()
{
   // freopen("in.txt","r",stdin) ;
    int n , m , k ;
    while(scanf("%d %d", &n , &m) != EOF)
    {
         if (n == 0 && m == 0)
             break ;

         int i ;
         for(i = 0 ; i < n ; i++)
         {
             F[i] = i ;
             num[i] = 1 ;
         }
        int u , v ;
        while(m--)
        {
            scanf("%d %d" , &k , &u) ;
            k-- ;
            while(k--)
            {
                scanf("%d" , &v) ;
                bing(u , v) ;
            }
        }
        printf("%d
" , num[find(0)]) ; //集合元素的个数保存在祖先结点的num数组里


    }
    return 0;
}