PAT_A1107#Social Clusters

Source:

PAT A1107 Social Clusters (30 分)

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 }

 

posted @ 2019-06-23 15:11  林東雨  阅读(222)  评论(0)  编辑  收藏  举报