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
4 3 1

题意：有n个同学，每个同学有k个兴趣爱好，如果两个同学兴趣爱好相同，则称他们为有联系。现给出n个同学，求可以构成的联系个数，以及每个联系网络拥有的人数

思路：运用并查集进行统计（初始化，寻找根节点，合并），注意联系是通过兴趣进行的，另开设一个数组存放爱好

#include<stdio.h>
#include<algorithm>
using namespace std;
const int maxn=1010;
int father[maxn];
bool isroot[maxn];//计数几个联系网络
int number[maxn]={0};
int course[maxn]={0};
bool cmp(int a,int b){//排序
	return a>b;
}
void init(int n){//初始化并查集
	for(int i=1;i<=n;i++){
		isroot[i]=true;//起始全部为独立网络
		father[i]=i;
	}
}
int find_father(int x){
	while(x!=father[x]){
		isroot[x]=false;//成为孩子
		x=father[x];
	}
	return x;
}
void Union(int a,int b){
	int fa=find_father(a);//合并网络
	int fb=find_father(b);
	if(fa!=fb){
		isroot[fb]=false;
		father[fb]=fa;
	}
}
int main(){
	int n,k;
	scanf("%d",&n);
  	init(n);
	for(int i=1;i<=n;i++){
		scanf("%d:",&k);
		for(int j=1;j<=k;j++){
			int m;
			scanf("%d",&m);
			if(course[m]==0){//i的喜好为m
				course[m]=i;
			}
			Union(i,course[m]);//合并，依据喜好合并
		}
	}
	int ans=0;
	for(int i=1;i<=n;i++){
		number[find_father(i)]++;
		if(isroot[i]==true){
			ans++;
		}
	}
	printf("%d\n",ans);
	sort(number+1,number+1+n,cmp);
	for(int i=1;i<=n;i++){
		if(number[i]!=0){
			ans--;
			if(ans>0){
				printf("%d ",number[i]);
			}else{
				printf("%d\n",number[i]);
			}
		}
	}	
}