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

题目描述：潜在的朋友圈

输入：第一行n，表示人数；接着n行，每行是第i个人选择的课程数及课程id号。

输出：第一行输出多少个圈，接着输出按照圈里人数的大小递减输出。

解题思路：并查集，求出有多少个朋友圈，按照朋友圈的人数从大到小输出每个圈的人数就行了。

#include<bits/stdc++.h>
using namespace std;
int n, m;
const int maxn = 2000;
vector<int> v[maxn];
int f[maxn], g[maxn];
set<int> s;
int find(int x){
	while(f[x]!=x){
		x = f[x];
	}
	return x;
}
void Union(int x, int y){
	x = find(x);
	y = find(y);
	f[x] = y;
}
bool cmp(int a, int b){
	return a > b;
}
int main()
{
	int k;
	scanf("%d", &n);
	for(int i=1; i<=n; i++) f[i] = i;
	for(int i=1; i<=n; i++){
		scanf("%d:", &m);
		for(int j=0; j<m; j++){
			scanf("%d", &k);
			v[k].push_back(i);
			s.insert(k);
		}
	}
	set<int>::iterator it;
	for(it=s.begin(); it!=s.end(); it++){//课程id 
		sort(v[*it].begin(),v[*it].end());
		int x, y;
		for(int i=0; i<v[*it].size(); i++){
			if(i==0) x = find(v[*it][i]);
			else{
				y = find(v[*it][i]);
				Union(x,y);
			}
		}
	}
	map<int,int> p;
	for(int i=1; i<=n; i++) p[find(i)]++;
	map<int,int>::iterator it1;
	int cnt=0; 
	for(it1=p.begin(); it1!=p.end(); it1++){
		g[cnt++] = it1->second;
	}
	sort(g,g+cnt,cmp);
	printf("%d\n", cnt);
	for(int i=0; i<cnt; i++){
		if(i) printf(" ");
		printf("%d", g[i]);
	}
	return 0;
}