1107. Social Clusters (30)
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: hi[1] hi[2] ... hi[Ki]
where Ki (>0) is the number of hobbies, and hi[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: 4Sample Output:
3 4 3 1
分析:这是一道并查集的题目,但是需要注意的是,我们集合的元素是人,而不是爱好,所以需要一个映射,将每一个爱好的编号映射成人的编号,再对人进行并的操作。另外,如何使用并查集求每个集合的元素,我们只需要新增加一个计数的数组,每次进行合并操作时,将对应的计数数组进行相加即可。
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> using namespace std; const int maxn=1100; const int INF=1e9; int father[maxn]; int num[maxn]; int hobby[maxn]; void init() { for(int i=0;i<maxn;i++) { father[i]=i; num[i]=1; hobby[i]=-1; } } int findFather(int x) { int a=x; while(x!=father[x]) x=father[x]; while(a!=father[a]) { int z=a; a=father[a]; father[z]=x; } return x; } int cnt; void uf(int a,int b) { int fa=findFather(a); int fb=findFather(b); if(fa!=fb) { father[fa]=fb; num[fb]+=num[fa]; cnt--; } } int n; bool cmp(int a,int b) { return a>b; } int main() { init(); cin>>n; cnt=n; for(int i=1;i<=n;i++) { int k; scanf("%d: ",&k); for(int j=0;j<k;j++) { int kj; cin>>kj; if(hobby[kj]==-1) { hobby[kj]=i; } else { uf(hobby[kj],i); } } } cout<<cnt<<endl; vector<int> ans; for(int i=1;i<=n;i++) { if(father[i]==i) ans.push_back(num[i]); } sort(ans.begin(),ans.end(),cmp); for(int i=0;i<ans.size();i++) { if(i>0) cout<<" "; cout<<ans[i]; } return 0; }