PAT甲级——1107 Social Clusters (并查集)

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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 (≤), 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

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。

 #include <iostream>
 #include <vector>
 #include <cmath>
 #include <algorithm>
 #define MaxNum 1001
 using namespace std;
 vector <int> S(MaxNum, );
 bool cmp(int a, int b) {
     return a > b;
 }
 int getK(char *c);
 void unionSet(int root, int X);
 int find(int X);
 int main()
 {
     int N, K;
     scanf("%d", &N);
     for (int i = ; i < N; i++) {
         int root, X;
         char c[];
         scanf("%s%d", c, &X);
         K = getK(c);
         root = find(X);
         S[root]--;//人数+1用负数表示
         for (int j = ; j < K; j++) {
             scanf("%d", &X);
             unionSet(root, X);
         }
     }
     vector <int> ans;
     for (int i = ; i < MaxNum; i++) {
         if (S[i] < )
             ans.push_back(abs(S[i]));
     }
     sort(ans.begin(), ans.end(), cmp);
     printf("%d\n", ans.size());
     printf("%d", ans[]);
     for (int i = ; i < ans.size(); i++)
         printf(" %d", ans[i]);
     printf("\n");
     return ;
 }
 int getK(char *c) {
     int sum = ;
     for (int i = ; c[i] != ':'; i++)
         sum = sum *  + c[i] - '';
     return sum;
 }
 void unionSet(int root, int X) {
     int rootX = find(X);
     if(S[rootX] <= && rootX != root){
         S[root] += S[rootX];
         S[rootX] = root;
     }
 }
 int find(int X) {
     if (S[X] <= )
         return X;
     else
         return S[X] = find(S[X]);//递归地压缩路径
 }