PAT A1107 Social Clusters (30分)

題目鏈接：https://pintia.cn/problem-sets/994805342720868352/problems/994805361586847744

題目描述
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.

輸入
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] … hi​​ [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].

輸出
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.

樣例輸入
8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4

樣例輸出
3
4 3 1

代碼

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> father, isRoot;
int cmp1(int a, int b){
    return a > b;
}
int findFather(int x) {
    int a = x;
    while(x != father[x])
        x = father[x];
    while(a != father[a]) {
        int z = a;
        a = father[a];
        father[z] = x;
    }
    return x;
}
void Union(int a, int b) {
    int faA = findFather(a);
    int faB = findFather(b);
    if(faA != faB)
        father[faA] = faB;
}
int main() {
    int n, k, t, cnt = 0;
    int course[1001] = {0};
    scanf("%d", &n);
    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++) {
        scanf("%d:", &k);
        for(int j = 0; j < k; j++) {
            scanf("%d", &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++;
    }
    printf("%d\n", cnt);
    sort(isRoot.begin(), isRoot.end(), cmp1);
    for(int i = 0; i < cnt; i++) {
        printf("%d", isRoot[i]);
        if(i != cnt - 1) printf(" ");
    }
    return 0;
}