PAT-A1107 Social Clusters（30 分）【并查集】

1107 Social Clusters（30 分）
When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A social cluster is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer N (≤1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:K​i​​: h​i​​[1] h​i​​[2] ... h​i​​[K​i​​]
where K​i​​ (>0) is the number of hobbies, and h​i​​[j] is the index of the j-th hobby, which is an integer in [1, 1000].
Output Specification:
For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4
Sample Output:
3

思路：
并查集
合并的是：喜欢同一个活动的人
查找的是：每个活动的人的根
集合：构建出喜欢用一个活动的不同集合

course[h]:用以记录任意一个喜欢活动h的人的编号
findFather（course[h]):这个人所在的社交网络的根结点

#include
using namespace std;

const int maxn=1005;
int father[maxn];
int isRoot[maxn];
int course[maxn];

int findFather(int x)
{
    int a = x;
    while(x!=father[x])
    {
        x=father[x];
    }
//路径压缩（可以不压缩）
    while(a!=father[a])
    {
        int z=a;
        a=father[a];
        father[z]=x;
    }
    return x;
}

void Union(int a,int b)
{
    int faA=findFather(a);
    int faB=findFather(b);
    if(faA!=faB)
    {
        father[faA]=faB;
    }
}

void init(int n)
{
    for(int i=1;i<=n;i++)
    {
        father[i]=i;
        isRoot[i]=false;
    }
}

bool cmp(int a,int b)
{
    return a>b;
}

int main()
{
    int n,k,h;
    scanf("%d",&n);
    init(n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d:",&k);
        for(int j=0;j