1107 Social Clusters (30分)图论求连通分量

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 : h​i​​ [1] h​i[2] … h​i[Ki]

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

题目大意:每个人都有爱好，如果两个人爱好相同就属于相同社交圈，求n个人一共形成多少个社交圈，并输出每个社交圈的人数

分析：将前面人的编号按照自己爱好下标存在vectorhobby【1000】中，之后的人都判断是否这个爱好下标对应的人数>0,如果大于0，说明这个爱好是公共爱好，接着就遍历这个向量vectorhobby[i],为这个爱好中的每个人建立一条边存放在vectorpersons【1000】中，之后就是求连通分量了。

代码:

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
vector<int>hobby[1001];
vector<int>person[1001];
vector<int>res;
bool flag[1001]={false};
void dfs(int root,int &sum){
	sum++;
	flag[root]=true;
	for(int i=0;i<person[root].size();++i){
		if(flag[person[root][i]]==false)
		dfs(person[root][i],sum);
	}
}
int cmp(int a,int b){
	return a>b;
}
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;++i){
		int k;
		scanf("%d: ",&k);
		for(int j=0,t;j<k;++j){
			scanf("%d",&t);
			if(hobby[t].size()>0)
			{
				for(int z=0;z<hobby[t].size();++z){
					int d=hobby[t][z];
					person[i].push_back(d);
					person[d].push_back(i);
				}
			}
			hobby[t].push_back(i);
		}
	}
	for(int i=1;i<=n;++i){
		int sum=0;
		if(flag[i]==false){
			dfs(i,sum);
			res.push_back(sum);
		}
		
	}
	sort(res.begin(),res.end(),cmp);
	cout<<res.size()<<endl;
	for(int i=0;i<res.size();++i){
		if(i)
		cout<<" ";
		cout<<res[i];
	}
	return 0;
}

运行截图: