PAT 1107 Social Clusters

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本文介绍了一种使用并查集算法解决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，算法能够有效地将具有共同兴趣的人群归为同一集群，并输出每个集群的人数。

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1107 Social Clusters (30)（30 分）
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:

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

分析
参考并查集简析

#include<iostream> //集合的查并补
#include<vector>
#include<algorithm>
using namespace std;
vector<int> father, isroot;
bool cmp(const int& a, const int& b){
  return a>b;
}
int findfather(int a){
  int b=a;
  while(father[a]!=a){
    a=father[a];
  }
  while(b != father[b]) { //压缩路径
        int z = b;
        b = father[b];
        father[z] = a;
    }
  return a;
}
void Union(int a, int b){ //并集
  int faA= findfather(a);
  int faB= findfather(b);
  if(faA!=faB) father[faA]=faB;
}
int main(){
  int N, cnt=0;
  cin>>N;
  vector<int> course(1001, 0);
  father.resize(N+1);
  isroot.resize(N+1);
  for(int i=1; i<=N; i++)
    father[i]=i;
  for(int i=1; i<=N; i++){
    int k;
    scanf("%d:",&k);
    for(int j=0; j<k; j++){
      int t;
      cin>>t;
      if(course[t]==0)
        course[t]=i;
      Union(i, findfather(course[t]));
    }
  }
  for(int i=1; i<=N; i++){
    isroot[findfather(i)]++;
  }
  for(int i=1; i<=N; i++){
    if(isroot[i]!=0)
      cnt++;
  }
  sort(isroot.begin(), isroot.end(), cmp);
  cout<<cnt<<endl;
  for(int i=0; i<cnt; i++)
    i==0?cout<<isroot[i]:cout<<" "<<isroot[i];
  return 0;
}