1047 Student List for Course (25 分)

2021/6/30 23:21:09

本文主要是介绍1047 Student List for Course (25 分),对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!

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;
}

 



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