PAT A1107 Social Clusters （30 分）——并查集

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​​[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].

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 <stdio.h>
 #include <algorithm>
 #include <set>
 #include <string.h>
 #include <vector>
 #include <math.h>
 #include <queue>
 using namespace std;
 const int maxn = ;
 int n;
 int father[maxn]={};
 int peo[maxn][maxn];
 set<int> hob;
 set<int> fah;
 int res[maxn]={};
 bool cmp(int a,int b){
     return a>b;
 }
 void init(){
     for(int i=;i<=maxn;i++){
         father[i]=i;
     }
 }
 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 uni(int a,int b){
     int fa=findfather(a);
     int fb = findfather(b);
     if(fa!=fb){
         father[fb]=fa;
     }
 }
 int main(){
     init();
     scanf("%d",&n);
     for(int i=;i<=n;i++){
         int k,fi;
         scanf("%d: %d",&k,&fi);
         peo[i][]=fi;
         hob.insert(fi);
         for(int j=;j<k;j++){
             int x;
             scanf("%d",&x);
             peo[i][j]=x;
             hob.insert(x);
             uni(fi,x);
         }
     }
     for(auto it:hob){
         fah.insert(findfather(it));
         //printf("%d %d\n",it,findfather(it));
     }
     printf("%d\n",fah.size());
     for(int i=;i<=n;i++){
         res[findfather(peo[i][])]++;
         //printf("%d %d\n",findfather(peo[i][0]),res[findfather(peo[i][0])]);
     }
     sort(res,res+maxn,cmp);
     for(int i=;i<fah.size();i++){
         printf("%d",res[i]);
         if(i<fah.size()-)printf(" ");
     }
 }

注意点：标准并查集，我是根据喜好来把人集合起来的，变量有点多，有点麻烦。看了大佬的代码，发现好像所有并查集的题目都是可以套模板的，他们是合并人，标记爱好，然后遍历isroot来得到集合个数