1107 Social Clusters (30 分)

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 i_n[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

感悟

不难的一道并查集,稍稍转换一下,将共同爱好集合并成一个根爱好,然后统计每个根爱好下面有多少人,然后排序一下就好.

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;
int a[N],used[N],root[N],sum[N];

int find(int x)
{
    return x == a[x] ? x : a[x] = find(a[x]);
}

void merge(int x, int y)
{
    a[find(y)] = find(x); 
}

int main()
{
    int n;
    
    for(int i = 0; i < N; i++) a[i] = i;
    
    scanf("%d",&n);
    for(int i = 1; i <= n; i++)
    {
        int k,x,y;
        scanf("%d: %d", &k, &x);
        x = find(x);
        for(int j = 1; j < k; j++)
        {
            scanf("%d",&y);
            merge(x,y);
        }
        root[i] = find(x);
    }
    for(int i = 1; i <= n; i++)
        sum[find(root[i])]++;
    sort(sum+1,sum+N,greater<int>());
    int cnt = 0;
    for(int i = 1 ; i <= n && sum[i]; i ++)cnt++;
    printf("%d\n",cnt);
    for(int i = 1 ; i <= n && sum[i]; i++)
        printf("%d%c", sum[i], sum[i+1] == 0 ? '\n' : ' ');

    return 0;
}