PAT 1039 Course List for Student (25分) 使用map(string, vector(int))

题目

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

题目大意

K个课程,给出每个课程的编号,选课人数,这些人的名字
N个学生来查询他们的选课列表,按输入顺序处理每个人的查询请求,输出他的姓名 选课数目 课程1编号 课程2编号 ...,要求课程编号输出顺序满足从小到大

思路分析

既然要得到每个学生的选课列表,而给出的学生姓名是字符串,要求输出的课程编号是整数,那就用一个 map<string, vector<int>>存储,其中,键是学生姓名,值是学生选课列表。

我们需要做的就是在读取输出过程中,转换输入(每个课程选课的有哪些人)转换成 学生以及对应的选课列表

这个很简单,就是“对号入座”。

比如对于输入中某一个课的情况,1号课,4个人选了,姓名分别如下

1 4
ANN0 BOB5 JAY9 LOR6

只需要这样简单处理

// 课号和选课人数
cin >> cno >> cnum;
// 选课的都是哪些人
while(cnum-- > 0) {
    // 选课人名字
    cin >> name;
    // 记录,这个人,选了这个课
    stu_cource[name].push_back(cno);
}

之后要查询某个学生的选课列表,只需要map[name]就能得到他的选课列表vector<int>,为了满足输出要求,对vector<int>进行一次sort()即可。

完成代码

#include<iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;

// 保存每个学生和他的选择列表
map<string, vector<int>> stu_cource;

int main() {
    // N个学生要查询,K个课
    int N, K;
    cin >> N >> K;
    // 每个课的编号,选课人数
    int cno, cnum;
    string name;
    while (K-- > 0) {
        // 课号和选课人数
        cin >> cno >> cnum;
        // 选课的都是哪些人
        while(cnum-- > 0) {
            // 选课人名字
            cin >> name;
            // 记录,这个人,选了这个课
            stu_cource[name].push_back(cno);
        }
    }
    // N个人查询
    while(N-- > 0) {
        cin >> name;
        // 得到他的选课列表
        vector<int> course = stu_cource[name];
        // 课号按顺序排
        sort(course.begin(), course.end());
        // 输出 姓名 几个课 课号1 课号2 ...
        cout << name << " " << course.size();
        for (int cno: course) 
            cout << " " << cno;
        cout << endl;
    }
    return 0;
}

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