【PAT甲级】Course List for Student

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本文链接：https://blog.csdn.net/weixin_42449444/article/details/89715339

Problem Description：

Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive integers: N (≤40,000), the number of students who look for their course lists, and K (≤2,500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students N​i​​ (≤200) are given in a line. Then in the next line, N​i​​ student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.

Output Specification:

For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student's name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.

Sample Input:

11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9

Sample Output:

ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0

解题思路：

我还是比较机智，这题直接把map和vector结合起来用就AC啦。建立一个map<string, vector<int> > m，m的key是学生的姓名，m的value是个int型的vector ，它用来存放该学生所选的所有课程的课程号。输入完成后无脑sort，然后输出每个学生的姓名name、学生选择的课程数m[name].size()，for-each输出该学生选择的所有课程的课程号即可。map和vector结合起来用很酷，难道不是吗？

AC代码：

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int N,K;  //N个学生、K门课程
    cin >> N >> K;
    map<string, vector<int> > m;
    while(K--)
    {
        int id,cnt;   //课程号id和选该门课程的学生数cnt
        cin >> id >> cnt;
        while(cnt--)
        {
            string name;
            cin >> name;
            m[name].push_back(id);
        }
    }
    while(N--)
    {
        string name;
        cin >> name;
        cout << name << " " << m[name].size();
        sort(m[name].begin(), m[name].end());  //升序排列学生选的课程号
        for(auto it : m[name])
        {
            cout << " " << it;
        }
        cout << endl;
    }
    return 0;
}