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 (≤), 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?? (>) is the number of hobbies, and [ 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
4 3 1

思路：

1. 在每次輸入個人的一個愛好後，都可能對如今的社交網絡產生變化，故都需要在此時進行合並操作
2. father數組在初始保存的根節點都是自己，表示自己是一個單獨的社交網絡
3. 這裏因為每個人可能有不止一個愛好，故還需輔助數組hobby，記錄任何一個有該愛好的人，之後再與當前讀入的人合並根節點即可

#include<cstdio>
#include<algorithm>
using
 
 namespace std;
const int maxn=1010;
int father[maxn]={0};
int isRoot[maxn]={0};
int hobby[maxn]={0};
int n,num,temp;

int findFather(int x){
    if(father[x]==x) return x;
    else{
        father[x]=findFather(father[x]);
        return father[x];
    }
}

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

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

int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        father[i]=i;
    for(int i=1;i<=n;i++){
        scanf("%d:",&num);
        for(int j=0;j<num;j++){
            scanf("%d",&temp);
            if(hobby[temp]==0)
                hobby[temp]=i;
            Union(i,hobby[temp]);
        }
    }
    for(int i=1;i<=n;i++){
        isRoot[findFather(i)]++;//此處不能用father[i]，可能存在嵌套關系 
    }
    int ans=0;
    for(int i=1;i<=n;i++)
        if(isRoot[i]!=0) ans++;
    printf("%d\n",ans);
    sort(isRoot+1,isRoot+n+1,cmp);
    for(int i=1;i<=ans;i++){
        printf("%d",isRoot[i]);
        if(i<ans) printf(" ");
    }
    return 0;
}

1107 Social Clusters （復雜並查集）