PAT 1107 Social Clusters

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:

K​i​​: h​i​​[1] h​i​​[2] ... h​i​​[K​i​​]

where K​i​​ (>) 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

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
vector<int> parent(1001), _count(1001), vis(1001, false);
int find(int x){
  int y=x, z;
  while(x!=parent[x]) x=parent[x];
  while(y!=x){
    z=parent[y];
    parent[y]=x;
    y=z;
  }
  return x;
}
 
bool _union(int a, int b){
  int roota=find(a), rootb=find(b);
  if(roota==rootb) return false;
  else{
      if(_count[roota]>_count[rootb]) {
          parent[rootb]=roota;
          _count[roota] += _count[rootb];
      }else{
          parent[roota]=rootb;
          _count[rootb] += _count[roota];
      }
    return true;
  }
}
//并查集解决这道题 很简单 用一个数组来记录根节点 用一个数组来记录每棵树的的人数  
//这一题的测试数据比较少  不压缩路径也没有关系
int main(){
  int n, i;
  scanf("%d", &n);
  fill(_count.begin(), _count.end(), 0);
  for(i=1; i<=1000; i++) parent[i]=i;//开始条件为i<1000有四个测试点不能通过  修改成i<=1000就能全部通过了
  for(i=0; i<n; i++){
    int cnt, j, temp, hobby;
    scanf("%d: %d", &cnt, &temp);
    _count[find(temp)]++;
    vis[temp]=true;
    for(j=1; j<cnt; j++){
      scanf("%d", &hobby);
      _union(temp, hobby);
      temp=hobby;
      vis[temp]=true;
    }
  }
  vector<int> ans;
  for(i=1; i<=1000; i++)
    if(parent[i]==i && vis[i]) ans.push_back(_count[i]);
  sort(ans.begin(), ans.end());
  printf("%d\n", ans.size());
  printf("%d", ans[ans.size()-1]);
  for(i=ans.size()-2; i>=0; i--) printf(" %d", ans[i]); 
  return 0;
}