1047 Student List for Course（25 分）

最新推荐文章于 2019-09-29 15:54:37 发布

原创最新推荐文章于 2019-09-29 15:54:37 发布 · 542 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲级专栏收录该内容

54 篇文章

订阅专栏

博客围绕浙江大学课程学生名单输出问题展开，给出输入输出规格及示例。输入包含学生总数、课程总数、学生选课信息，输出要求按课程号升序输出各课程学生名单。解题思路是用string型vector数组记录学生名字，输出前排序，且用cout最后一点会超时。

1047 Student List for Course（25 分）
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 (≤40,000), the total number of students, and K (≤2,500), 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

解题思路：
用string型的vector数组记录每个课程中的学生名字，输出前对其排序。要注意用cout的话最后一个点会超时。

#include<cstdio> 
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn=2550;
vector<string>course[maxn];

int main(){
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++){
        string name;
        int c,x;
        cin>>name>>c;
        for(int j=0;j<c;j++){
            scanf("%d",&x);
            course[x].push_back(name) ;
        }
    }
    for(int i=1;i<=k;i++){
        printf("%d %d\n",i,course[i].size() );
        sort(course[i].begin() ,course[i].end() );
        for(int j=0;j<course[i].size() ;j++){
            printf("%s\n",course[i][j].c_str());
        }
    }
    return 0;
}