A1107. Social Clusters

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<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int cnt[1001] = {0,0}, father[1001], hobby[1001] = {0,0};
int N;
int findFather(int x){
    int a = x;
    while(father[x] != x){
        x = father[x];
    }
    while(a != x){
        int temp = a[father];
        a[father] = x;
        a = temp;
    }
    return x;
}
void union_(int a, int b){   //a集合的根节点不变
    int tempa = findFather(a);
    int tempb = findFather(b);
    if(tempa != tempb){
        father[tempb] = a;
    }
}
bool cmp(int a, int b){
    return a > b;
}
int main(){
    int N;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++){
        father[i] = i;
    }
    for(int i = 1; i <= N; i++){
        int num, temp;
        scanf("%d:", &num);
        for(int j = 0; j < num; j++){
            scanf("%d", &temp);
            if(hobby[temp] == 0){
                hobby[temp] = i;
            }else{
                union_(hobby[temp], i);
            }
        }
    }
    for(int i = 1; i <= N; i++){
        int root = findFather(i);
        cnt[root]++;
    }
    int ans = 0;
    for(int i = 1; i <= N; i++){
        if(cnt[i] != 0)
            ans++;
    }
    sort(cnt, cnt + N + 1, cmp);
    printf("%d\n%d", ans, cnt[0]);
    int k = 1;
    while(k < N && cnt[k] != 0){
        printf(" %d", cnt[k]);
        k++;
    }
    cin >> N;
    return 0;
}

总结：

1、题意：看了书上的解释才搞明白这道题什么意思。并不是说一个set里的所有人都要有共同的hobby，而是当两个人有至少一个共同hobby时，他们就处于一个set。比如A的hobby是1、2，B的hobby是2、3，则AB在一个set中。而C的hobby是3、4，则B、C在一个set中。AB同set，BC同set，则ABC同属一个set。其实就是考并查集。

2、并查集的要点：

• father[a] = b，表示a的父节点是b。若father[i] = i，则i是根节点。 初始化时，每个节点都初始化为根节点。
• 查找根节点：当x != father[x] 时，不断进行x = father[x] 的操作即可。
• 合并：为防止出现环路，只能对不同的集合做合并。所以合并a、b时，先找到a的根节点roota， b的根节点rootb，如果roota ！= rootb， 则 father[rootb] = roota。严格按这个流程做可以避免出错。不要仅仅把b的父亲设为roota。
• 路径压缩：为了降低查找的复杂度。可以放在查找根节点的函数中，当找到 x 的根节点 root 时，再从 x 往根节点回溯一次，沿途所有节点的父节点均设置为 root。

3、只有两个人有共同hobby才将他们所在的两个set做合并，但如果对每个人保存一个hobby列表，然后每两个人检查是否有共同爱好会很费时。可以设置一个hobby数组，hobby[i]仅仅记录一个人，就是首个读入的有这个爱好的人，并将这个人作为他所在set的root节点一直不变。