PAT(并查集)——1107. Social Clusters (30)

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:
Ki: hi[1] hi[2] … hi[Ki]
where Ki (>0) is the number of hobbies, and hi[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

题目大意：

两个人有一样的爱好就称为处于同一个社交网络，求一共有多少个社交网络，人数从大到小输出。

题目解析：

以人作为节点，对爱好做一个到人的映射，其他的就跟正常的并查集题目一样了。

具体代码：

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
#define MAXN 1005
int parent[MAXN];
int num[MAXN];
int course[MAXN];
void init(){
	for(int i=0;i<MAXN;i++){
		parent[i]=i;
		course[i]=0;
		num[i]=0;
	}
}
int find_parent(int x){
	int a=x;
	while(x!=parent[x])
		x=parent[x];
	while(a!=parent[a]){//路径压缩 
		int b=a;
		parent[b]=x;
		a=parent[a];
	}
	return x;
}
void Union(int a,int b){
	int fa=find_parent(a);
	int fb=find_parent(b);
	if(fa!=fb){
		parent[fa]=fb;
	}
}
bool cmp(int a,int b){
	return a>b;
} 
int main()
{
    int n;
    cin>>n;
    init();
	for(int i=1;i<=n;i++){
		int k;
		scanf("%d:",&k);
		while(k--){
			int tmp;
			cin>>tmp;
			if(course[tmp]==0)
				course[tmp]=i;
			Union(i,find_parent(course[tmp]));
		}
	}
	for(int i=1;i<=n;i++){
		num[find_parent(i)]++;
	}
		
	sort(num,num+MAXN,cmp);
	vector<int> v;
	for(int i=0;i<MAXN;i++){
		if(num[i]==0)
			break;
		v.push_back(num[i]);
	}
	cout<<v.size()<<endl;
	for(int i=0;i<v.size();i++){
		if(i==v.size()-1)
			cout<<v[i];
		else
			cout<<v[i]<<" ";
	}
    return 0;
}