Source:
Description:
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 (≤), 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?? (>) is the number of hobbies, and [ 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
Keys:
Code:
1 /*
2 time: 2019-06-23 14:07:12
3 problem: PAT_A1107#Social Clusters
4 AC: 34：25
5
6 题目大意：
7 把一群具有相同爱好的人归为一个社交圈，找出所有的社交圈
8 输入：
9 第一行给出，总人数N<=1e3，编号从1~N
10 接下来N行，给出第i个人的，爱好总数K，各个爱好
11 输出：
12 第一行给出，社交圈总数
13 第二行给出，各个社交圈的人数，从多到少
14
15 基本思路：
16 基于兴趣做并查集操作，
17 输入每个人的兴趣，首个兴趣的Hash值+1，标记人数
18 统计父结点个数及其孩子的哈希值即可
19 */
20 #include<cstdio>
21 #include<set>
22 #include<algorithm>
23 using namespace std;
24 const int M=1e3+10;
25 int fa[M],man[M]={0},ans[M]={0};
26
27 int Father(int v)
28 {
29 int x=v,s;
30 while(fa[v] != v)
31 v = fa[v];
32 while(fa[x] != x){
33 s = fa[x];
34 fa[x] = v;
35 x = s;
36 }
37 return v;
38 }
39
40 void Union(int v1, int v2)
41 {
42 int f1 = Father(v1);
43 int f2 = Father(v2);
44 fa[f2] = f1;
45 Father(v2);
46 }
47
48 int main()
49 {
50 #ifdef ONLINE_JUDGE
51 #else
52 freopen("Test.txt", "r", stdin);
53 #endif // ONLINE_JUDGE
54
55 for(int i=0; i<M; i++)
56 fa[i]=i;
57
58 int n,m,h1,h2;
59 set<int> hobby,clster;
60 scanf("%d", &n);
61 for(int i=0; i<n; i++)
62 {
63 scanf("%d:%d", &m,&h1);
64 man[h1]++;
65 hobby.insert(h1);
66 for(int j=1; j<m; j++)
67 {
68 scanf("%d", &h2);
69 hobby.insert(h2);
70 Union(h1,h2);
71 h1=h2;
72 }
73 }
74 for(auto it=hobby.begin(); it!=hobby.end(); it++){
75 ans[Father(*it)] += man[*it];
76 clster.insert(Father(*it));
77 }
78 printf("%d\n", clster.size());
79 sort(ans, ans+M, greater<int>() );
80 for(int i=0; i<clster.size(); i++)
81 printf("%d%c", ans[i], i+1==clster.size()?‘\n‘:‘ ‘);
82
83 return 0;
84 }
原文地址：https://www.cnblogs.com/blue-lin/p/11072913.html