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[1005], father[1005];
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 = 0; i <= n; ++i)father[i] = i;//初始化
    for (int i = 1; i <= n; ++i) 
    {
        scanf("%d:", &m);
        while (m--) 
        {
            cin >> a;
            if (hobby[a] == 0)//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 + 1, 0);//储存每个集合的人数
    for (int i = 1; i < n + 1; ++i)
        ++result[findFather(i)];//向前寻找father
    sort(result.begin(), result.end(), [](int a, int b) {return a > b; });
    int cnt = 0;
    for (auto t : result)
        if (t != 0)
            cnt++;
    cout << cnt << endl;
    for (int i = 0; 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 > 0 ? " " : "", result[i]);
    return 0;
}