PAT甲级——1107 Social Clusters (并查集)

本文同步发布在CSDN：https://blog.csdn.net/weixin_44385565/article/details/90409731

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

where

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:

Sample Output:

3
4 3 1

题目大意：将N个人按照兴趣爱好分类，爱好有交集的归为一个社交团体，求社交团体的个数，然后将团体里面的人数进行降序输出。

思路：在并查集的操作上稍作修改，用根节点的值进行人数的统计，以每一行的第一个爱好代号作为当前数据的root，将之后的X与root进行union操作，若S[rootX]不等于root且 ≤ 0，意味着当前团体已经有了 | S[rootX] | 人，将S[rootX]的值加到S[root]上再进行合并。注意每一行开头的K:需要用字符串数组处理来获取K的值。

 1 #include <iostream>
 2 #include <vector>
 3 #include <cmath>
 4 #include <algorithm>
 5 #define MaxNum 1001
 6 using namespace std;
 7 vector <int> S(MaxNum, 0);
 8 bool cmp(int a, int b) {
 9     return a > b;
10 }
11 int getK(char *c);
12 void unionSet(int root, int X);
13 int find(int X);
14 int main()
15 {
16     int N, K;
17     scanf("%d", &N);
18     for (int i = 0; i < N; i++) {
19         int root, X;
20         char c[5];
21         scanf("%s%d", c, &X);
22         K = getK(c);
23         root = find(X);
24         S[root]--;//人数+1用负数表示
25         for (int j = 1; j < K; j++) {
26             scanf("%d", &X);
27             unionSet(root, X);    
28         }
29     }
30     vector <int> ans;
31     for (int i = 1; i < MaxNum; i++) {
32         if (S[i] < 0)
33             ans.push_back(abs(S[i]));
34     }
35     sort(ans.begin(), ans.end(), cmp);
36     printf("%d
", ans.size());
37     printf("%d", ans[0]);
38     for (int i = 1; i < ans.size(); i++)
39         printf(" %d", ans[i]);
40     printf("
");
41     return 0;
42 }
43 int getK(char *c) {
44     int sum = 0;
45     for (int i = 0; c[i] != ':'; i++)
46         sum = sum * 10 + c[i] - '0';
47     return sum;
48 }
49 void unionSet(int root, int X) {
50     int rootX = find(X);
51     if(S[rootX] <=0 && rootX != root){
52         S[root] += S[rootX];
53         S[rootX] = root;
54     }
55 }
56 int find(int X) {
57     if (S[X] <= 0)
58         return X;
59     else
60         return S[X] = find(S[X]);//递归地压缩路径
61 }