PAT-A1107 Social Clusters 题目内容及题解

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

题目大意

题目给出用户列表，所有有共同爱好的人可以组成一个潜在的社交团体，要求输出用户列表中这样的社交团体的组合人数和每个组合的人数。

解题思路

1. 读入用户列表，读入过程中将爱好用并查集归类；
2. 计数每个并查集的元素数；
3. 输出结果并返回零值。

代码

#include<cstdio>
#include<algorithm>
using namespace std;
#define maxn 1010

int father[maxn],isRoot[maxn];
int course[maxn],N,cnt=0;

bool cmp(int a,int b){
    return a>b;
}

int FindFather(int x) {
    int a=x,z;
    while(x!=father[x]){
        x=father[x];
    }
    while(a!=father[a]){
        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 i,j,k,a;
    scanf("%d",&N);
    for(i=1;i<=N;i++){
        father[i]=i;
        scanf("%d:",&k);
        for(j=0;j<k;j++){
            scanf("%d",&a);
            if(course[a]==0){
                course[a]=i;
            }
            Union(i,FindFather(course[a]));
        }
    }
    for(i=1;i<=N;i++){
        isRoot[FindFather(i)]++;
    }
    for(i=1;i<=N;i++){
        if(isRoot[i]!=0){
            cnt++;
        }
    }
    sort(isRoot,isRoot+N+1,cmp);
    printf("%d\n",cnt);
    for(i=0;i<cnt;i++){
        printf("%d",isRoot[i]);
        if(i<cnt-1){
            printf(" ");
        }else{
            printf("\n");
        }
    }
    return 0;
}

运行结果