1047. Student List for Course (25)

深入探讨信息技术领域的关键技术与应用

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#PAT #算法

本文详细介绍了信息技术领域的核心技术，包括但不限于前端开发、后端开发、移动开发、游戏开发、大数据开发等细分领域，旨在为读者提供全面的技术洞察与实践指导。从基础知识到高级应用，涵盖多种开发工具、开发环境、测试方法、运维策略及最新的技术趋势。

Zhejiang University has 40000 students and provides 2500 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 (<=40000), the total number of students, and K (<=2500), 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

思路：建立2500个课程的vector，存放学生的姓名（最开始的时候用了list，超时；不要用c++的cin和cout）

#include <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"vector"
#include"string"
#include"algorithm"


using namespace std;


int main()
{
    long N;
    int K,C;
    cin>>N>>K;
     scanf("%ld%d",&N,&N);


    int order;
    vector<string> l[2501];


    string name;
    for(int i=1;i<=N;i++){
        cin>>name;
        scanf("%d",&C);
        for(int j=1;j<=C;j++){
            scanf("%d",&order);
            l[order].push_back(name);
        }
    }

    for(int i=1;i<=K;i++){
        int size = l[i].size();
        printf("%d %d\n",i,size);
        if(size)
        {
            sort(l[i].begin(),l[i].end());


            vector<string>::iterator iter;
            for(iter=l[i].begin();
                iter<l[i].end();
                iter++){
                printf("%s\n",(*iter).c_str());
            }
        }
    }
    return 0;
}