PAT 甲级A1107 Social Clusters（并查集）

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805361586847744
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​​ [Ki​]
where K​i​​ (>0) is the number of hobbies, and hi​​ [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

题目类型：并查集。
大概翻译：有n个人，每人喜欢k个活动，若两人喜欢的活动存在相同的一项则他们属于同一个社交网络，求一共形成了多少社交网络。
思路：先将getRoot()、Merge()函数写好，getRoot函数用于获取祖先节点，Merge函数用于压缩路径。在每次读如具体的喜好时都将先寻找所在树，进行并树；最后统计isRoot数组内不为0的个数即为一共多少棵树（isRoot内每个不为0的数即为每棵树所含的节点数即有多少个人在同一个社交网络），将isRoot排序输出。

源代码：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
vector<int> parent, isRoot; //父母数组、树的个数
int course[1010]; //course[t]表示任意⼀一个喜欢t活动的⼈人的编号；
int cmp(int a, int b) {return a > b;}
int getRoot(int x) { //得到祖先节点
    int a = x;
    while(x != parent[x])
        x = parent[x];
  //下面一段参考与柳婼的代码，用于优化时间，否则会有测试用例超时
    while(a != parent[a]) { 
        int z = a;
        a = parent[a];
        parent[z] = x;
    }
    return x;
}
void Merge(int a, int b) { //合并a， b所属的树
    int pa = getRoot(a);
    int pb = getRoot(b);
    if (pa != pb)   parent[pa] = pb; //将a树并到b树下面。
}
int main() {
    int n, k, t, cnt = 0;
    scanf("%d", &n); //n个人
    parent.resize(n + 1); //resize函数用于重置vector数组的大小
    isRoot.resize(n + 1);
    //在输入之前，每个节点都可以看成祖先是自己的节点
    for (int i = 1; i <= n; i++)    parent[i] = i; 
    for (int i = 1; i <= n; i++) {
        scanf("%d:", &k); //所喜欢的k个活动
        for (int j = 0; j < k; j++) {
            scanf("%d", &t); //具体活动
 //如果当前的课程t，之前并没有⼈人喜欢过，那么就course[t] = i，i为它自己的编号，表示i为喜欢course[t]的⼀个⼈的编号。
            if (course[t] == 0)    course[t] = i; 
            Merge(i, getRoot(course[t])); //得到合并后的树
        }
    }
   //统计树的个数，即为一共分为多少网络
    for (int i = 1; i <= n; i++)    isRoot[getRoot(i)]++;
    for(int i = 1; i <= n; i++)    if(isRoot[i] != 0) cnt++; 
    printf("%d\n", cnt);
    //对已经统计好的数据排序，sort函数属于algorithm库
    sort(isRoot.begin(), isRoot.end(), cmp);
    for(int i = 0; i < cnt; i++) {
        printf("%d", isRoot[i]);
        if(i != cnt - 1) printf(" ");
    }
    return 0;
}

（代码部份思路参照柳神写出，在这膜一下！）