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 = 1011;
int n;
int father[maxn]={0};
int peo[maxn][maxn];
set<int> hob;
set<int> fah;
int res[maxn]={0};
bool cmp(int a,int b){
    return a>b;
}
void init(){
    for(int i=1;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=1;i<=n;i++){
        int k,fi;
        scanf("%d: %d",&k,&fi);
        peo[i][0]=fi;
        hob.insert(fi);
        for(int j=1;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
",it,findfather(it));
    }
    printf("%d
",fah.size());
    for(int i=1;i<=n;i++){
        res[findfather(peo[i][0])]++;
        //printf("%d %d
",findfather(peo[i][0]),res[findfather(peo[i][0])]);
    }
    sort(res,res+maxn,cmp);
    for(int i=0;i<fah.size();i++){
        printf("%d",res[i]);
        if(i<fah.size()-1)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