PAT 甲级 1107 Social Clusters (30分)(并查集)
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 (≤), 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 (>) is the number of hobbies, and [ 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
题意:
有n个人,每个人喜欢k个活动,如果两个人有任意一个活动相同,就称为他们处于同一个社交网络。求这n个人一共形成了多少个社交网络。
题解:
并查集。通过vector是按兴趣爱好分类,再遍历vector将同一爱好的人合并。最后用a数组统计聚类个数及类别里的个数。
AC代码:
#include<bits/stdc++.h> using namespace std; vector<int>v[1005]; int n; int a[1005]; int p=0; int fa[1005]; int find(int x){ if(x==fa[x]) return x; return fa[x]=find(fa[x]); } void unio(int x,int y){ int yy=find(y); int xx=find(x); if(xx!=yy){ fa[yy]=xx; } } bool cmp(int x,int y){ return x>y; } int main(){ cin>>n; for(int i=1;i<=n;i++){ fa[i]=i; char s; int m; cin>>m>>s; for(int j=1;j<=m;j++){ int x; cin>>x; v[x].push_back(i); } } memset(a,0,sizeof(a)); for(int i=1;i<=1000;i++){ if(v[i].size()>0){ int x=v[i].at(0); for(int j=1;j<v[i].size();j++){ unio(x,v[i].at(j));//相同爱好合并 } } } for(int i=1;i<=n;i++){//统计聚类个数 a[find(i)]++; } sort(a+1,a+1+1000,cmp); for(int i=1;i<=1000;i++){ if(a[i]>0){ p++; } } cout<<p<<endl; for(int i=1;i<=p;i++){ cout<<a[i]; if(i!=p) cout<<" "; } return 0; }