1107 Social Clusters （30 point(s)）

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

经验总结：

emmmm 经典的并查集题目，这里要注意一点对于所有的人，都要进行两两判定是否有相同的爱好（除了自己和自己），如果有相同的爱好，就要合并，关于并查集的部分，是最基础的并查集的合并与查找，最后再统计群体的个数以及各个群体内部的人数就行了~

AC代码

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
const int maxn=1010;
vector<int> Node[maxn],ans;
int n,k,father[maxn],num[maxn]={0};
bool judge(int a,int b)
{
	if(Node[a].size()>Node[b].size())
	{
		int temp=a;
		a=b;
		b=temp;
	}
	for(int i=0;i<Node[a].size();++i)
	{
		vector<int>::iterator res=find(Node[b].begin(),Node[b].end(),Node[a][i]);
		if(res!=Node[b].end())
			return true;
	}
	return false;
}
bool cmp(int a,int b)
{
	return a>b;
}
int findFather(int x)
{
	int a=x;
	while(x!=father[x])
		x=father[x];
	while(a!=father[a])
	{
		int z=father[a];
		father[a]=x;
		a=z;
	}
	return x;
}
void Union(int a,int b)
{
	int x=findFather(a);
	int y=findFather(b);
	if(x!=y)
		father[x]=y;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	{
		father[i]=i;
		scanf("%d: ",&k);
		Node[i].resize(k);
		for(int j=0;j<k;++j)
			scanf("%d",&Node[i][j]);
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j)
			if(i!=j&&judge(i,j))
				Union(i,j);
	for(int i=1;i<=n;++i)
		++num[findFather(i)];
	for(int i=1;i<=n;++i)
		if(num[i]!=0)
			ans.push_back(num[i]);
	sort(ans.begin(),ans.end(),cmp);
	printf("%d\n",ans.size());
	for(int i=0;i<ans.size();++i)
		printf("%d%c",ans[i],i<ans.size()-1?' ':'\n');
	return 0;
}