PAT甲级——A1107 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 <numeric>
 #include <vector>
 #include <algorithm>
 using namespace std;
 int hobby[], father[];
 int findFather(int x)
 {//查找父亲结点并进行路径压缩
     if (x == father[x])
         return x;
     int temp = findFather(father[x]);
     father[x] = temp;
     return temp;
 }
 void unionSet(int a, int b)
 {//合并两个集合
     int ua = findFather(a), ub = findFather(b);
     if (ua != ub)
         father[ua] = ub;//这里是关键，即将此位置的father改为最近有共同爱好的人
 }
 int main() {
     int n, m, a;
     cin >> n;
     for (int i = ; i <= n; ++i)father[i] = i;//初始化
     for (int i = ; i <= n; ++i)
     {
         scanf("%d:", &m);
         while (m--)
         {
             cin >> a;
             if (hobby[a] == )//没有人有当前这个爱好
                 hobby[a] = i;//i作为第一个有该爱好的人
             else//有人喜欢该爱好
                 unionSet(hobby[a], i);//将有同样爱好的两个人合并为一个集合
         }
     }
     vector<int>result(n + , );//储存每个集合的人数
     for (int i = ; i < n + ; ++i)
         ++result[findFather(i)];//向前寻找father
     sort(result.begin(), result.end(), [](int a, int b) {return a > b; });
     int cnt = ;
     for (auto t : result)
         if (t != )
             cnt++;
     cout << cnt << endl;
     for (int i = ; i < cnt; ++i)//输出result前cnt个元素(result已经从大到小排序，输出的都是集合个数不为0的)
         printf("%s%d", i >  ? " " : "", result[i]);
     return ;
 }