1047 Student List for Course

Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤40,000), the total number of students, and K (≤2,500), the total number of courses. Then N lines follow, each contains a student’s name (3 capital English letters plus a one-digit number), a positive number C (≤20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students’ names in alphabetical order. Each name occupies a line.

Sample Input:

10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5

Sample Output:

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

题意

给定学生的选课信息，需要输出每个课程选课的学生数量以及学生姓名。为了方便，所有课程编号在 1 ~ k 之间。另外，输出课程选课学生的姓名时按照字典序输出。

思路

具体思路如下：

输入每个学生的选课信息，用一个存储 vector 的数组来保存每个课程选课的学生姓名。

遍历数组中的每个课程，先输出课程号和该课程的选课人数，再对该课程的 vector 进行排序，最后再依次输出每个学生的姓名。

代码

#include<bits/stdc++.h>
using namespace std;
const int N = 2510;
vector<string> ans[N];
int main()
{
    int n, k;
    cin >> n >> k;
    //输入每个学生的选课信息
    for (int i = 1; i <= n; i++)
    {
        string name;
        int m;
        cin >> name >> m;
        while (m--)
        {
            int c;
            cin >> c;
            ans[c].push_back(name);
        }
    }
    //输出每个课程选课的学生
    for (int i = 1; i <= k; i++)
    {
        cout << i << " " << ans[i].size() << endl;
        sort(ans[i].begin(), ans[i].end());
        for (auto& s : ans[i])
            cout << s << endl;
    }
    return 0;
}