zoukankan      html  css  js  c++  java
  • 1047. Student List for Course (25)

    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
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<vector>
    using namespace std;
    
    char name[40010][5];
    vector<int> course[2510];
    
    bool cmp(int a,int b){
        return strcmp(name[a],name[b]) < 0;
    }
    
    int main(){
    
        int n,k;
        scanf("%d%d",&n,&k);
        int courseID,c;
        for(int i = 0; i < n; i++){
            scanf("%s %d",name[i],&c);
        
            for(int j = 0; j < c; j++){
                scanf("%d",&courseID);
                course[courseID].push_back(i);
            }
        }
        for(int i = 1; i <= k; i++){
            printf("%d %d
    ",i,course[i].size());
            sort(course[i].begin(),course[i].end(),cmp);
            for(int j = 0; j < course[i].size(); j++){
                printf("%s
    ",name[course[i][j]]);
            }
        } 
        return 0;
    }
  • 相关阅读:
    太精辟了!从学校到职场的十条经典语录!
    平庸领导下跳棋,伟大领导下象棋(转)
    新官上任前的十一大基本功
    病母私自出房有感
    你为何还会出现在我梦里
    创业辛酸
    Goldengate can't extract data from compressed table
    配置GoldenGate同步DDL语句(3)
    Goldengate各build与Oracle数据库版本间的兼容性
    11g新特性:Note raised when explain plan for create index
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/8909103.html
Copyright © 2011-2022 走看看