1107. Social Clusters

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本文介绍了一个社交网络中寻找具有相同兴趣爱好的人群集群的问题。通过输入每个人的兴趣列表，使用并查集算法找出所有集群，并输出每个集群的人数。

1107. Social Clusters (30)

时间限制

1000 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

Sample Output:

3
4 3 1

并查集问题。

#include<vector>
#include<iostream>
#include<algorithm>

using namespace std;

typedef struct node
{
	bool f;
	int root;
	vector<int>c;
};

int p[1001];
node h[1001];

int findr(int num)
{
	if(h[num].root==num)
		return num;
	int r=findr(h[num].root);
	h[num].root=r;
	return r;
}
int main()
{
	int n;
	cin>>n;
	int ord[1001];
	for(int i=0;i<=1000;i++)
	{
		h[i].root=i;
		h[i].f=false;
		ord[i]=0;
	}
	for(int i=1;i<=n;i++)
	{
		int ph;
		scanf("%d:",&ph);
		int hid=0;
		for(int j=0;j<ph;j++)
		{
			cin>>hid;
			h[hid].f=true;
			h[hid].c.push_back(i);
		}
		p[i]=hid;
	}
	for(int i=1;i<=1000;i++)
	{
		if(h[i].f)
		{
			int r1=findr(h[i].root);
			for(int j=0;j<h[i].c.size();j++)
			{
				int r2=findr(p[h[i].c[j]]);
				if(r1!=r2)
				{
					h[r2].root=r1;
				}
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		p[i]=findr(p[i]);
		ord[p[i]]++;
	}
	
	int cnt=0;
	for(int i=1;i<=1000;i++)
		if(h[i].f&&h[i].root==i)
			cnt++;
	cout<<cnt<<endl;
	sort(ord,ord+1001);
	cout<<ord[1000];
	for(int i=999;i>=0;i--)
	{
		if(ord[i]==0)
			break;
		cout<<" "<<ord[i];
	}
	cout<<endl;
	return 0;
}