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