PAT_A1107#Social Clusters

Source:

PAT A1107 Social Clusters (30 分)

Description:

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

Keys:

• 并查集（Union-Find）

Code:

 /*
 time: 2019-06-23 14:07:12
 problem: PAT_A1107#Social Clusters
 AC: 34：25

 题目大意：
 把一群具有相同爱好的人归为一个社交圈，找出所有的社交圈
 输入：
 第一行给出，总人数N<=1e3，编号从1~N
 接下来N行，给出第i个人的，爱好总数K，各个爱好
 输出：
 第一行给出，社交圈总数
 第二行给出，各个社交圈的人数，从多到少

 基本思路：
 基于兴趣做并查集操作，
 输入每个人的兴趣，首个兴趣的Hash值+1，标记人数
 统计父结点个数及其孩子的哈希值即可
 */
 #include<cstdio>
 #include<set>
 #include<algorithm>
 using namespace std;
 const int M=1e3+;
 int fa[M],man[M]={},ans[M]={};

 int Father(int v)
 {
     int x=v,s;
     while(fa[v] != v)
         v = fa[v];
     while(fa[x] != x){
         s = fa[x];
         fa[x] = v;
         x = s;
     }
     return v;
 }

 void Union(int v1, int v2)
 {
     int f1 = Father(v1);
     int f2 = Father(v2);
     fa[f2] = f1;
     Father(v2);
 }

 int main()
 {
 #ifdef ONLINE_JUDGE
 #else
     freopen("Test.txt", "r", stdin);
 #endif // ONLINE_JUDGE

     for(int i=; i<M; i++)
         fa[i]=i;

     int n,m,h1,h2;
     set<int> hobby,clster;
     scanf("%d", &n);
     for(int i=; i<n; i++)
     {
         scanf("%d:%d", &m,&h1);
         man[h1]++;
         hobby.insert(h1);
         for(int j=; j<m; j++)
         {
             scanf("%d", &h2);
             hobby.insert(h2);
             Union(h1,h2);
             h1=h2;
         }
     }
     for(auto it=hobby.begin(); it!=hobby.end(); it++){
         ans[Father(*it)] += man[*it];
         clster.insert(Father(*it));
     }
     printf("%d\n", clster.size());
     sort(ans, ans+M, greater<int>() );
     for(int i=; i<clster.size(); i++)
         printf("%d%c", ans[i], i+==clster.size()?'\n':' ');

     return ;
 }