标签：25 1047 course LOR6 List number courses each BOB5

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 (≤), the total number of students, and K (≤), 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 (≤) 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：为每一个课程建立一个vector数组，存储选择该门课的学生

#include<bits/stdc++.h>
using namespace std;
const int maxn=1010;
vector<vector<string> > ve;
bool cmp(string a,string b){
    return a<b;
}
int main(){
    int m,n;
    scanf("%d %d",&m,&n);
    ve.resize(m);
    for(int i=0;i<m;i++){
        string st;
        int rig,stu;
        cin>>st;
        scanf("%d",&rig);
        for(int j=0;j<rig;j++){
            scanf("%d",&stu);
            ve[stu-1].push_back(st);
        }
    }
    for(int i=0;i<n;i++){
        printf("%d %d\n",i+1,ve[i].size());
        sort(ve[i].begin(),ve[i].end(),cmp);
        for(int j=0;j<ve[i].size();j++){
            printf("%s\n",ve[i][j].c_str());
        }
    }
    return 0;
}

 

标签：25,1047,course,LOR6,List,number,courses,each,BOB5
来源： https://www.cnblogs.com/dreamzj/p/14956812.html

本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享；
2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关；
3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关；
4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除；
5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。