1047. Student List for Course (25)-蒲公英云

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

题目大意：

浙江大学有40000名学生，提供2500门课程。现在根据每个学生的课程注册列表，你应该输出所有课程的学生名单。
输入规格：
每个输入文件包含一个测试用例。对于每个测试用例，第一行包含两个数字：N(<=40000)，是学生的总数，K（<=2500）,是课程总数。然后是N行，每行包含一个学生的名字（3个大写英文字母加上一个1位数字），一个正整数C（<=20），这个是学生注册的课程的数量，然后是C个课程的编号。为了简单起见，课程编号从1到K。
输出规范：
对于每个测试用例，一课程编号的递增顺序打印所有课程的的学生姓名列表。对于每一门课程，首先在一行上打印出课程编号和和注册学生的数量，用空格隔开。然后按字母顺序输出学生的名字。每个名字占一行。

代码：

#include<stdio.h>
#include<algorithm>
#include<vector>
#include<string.h>
using namespace std;
struct node
{
    char name[5];
}stud[180000];
vector<int> course[2501];
bool cmp(int a,int b)
{
    return a<b;
}
int change(char a[])
{
    return (a[0]-'A')*26*26*10+(a[1]-'A')*26*10+(a[2]-'A')*10+a[3]-'0';
}
int main()
{
    int i,j,n,m,k,t,id;
    char name[5];
    scanf("%d %d",&n,&m);
    for(i=0;i<n;i++)
    {
        scanf("%s %d",name,&k);
        id=change(name);
        strcpy(stud[id].name,name);
        for(j=0;j<k;j++)
        {
            scanf("%d",&t);
            course[t].push_back(id);
        }
    }
    for(i=1;i<=m;i++)
    {
       printf("%d %d\n",i,course[i].size());
       sort(course[i].begin(),course[i].end(),cmp);
       for(j=0;j<course[i].size();j++)
       {
           printf("%s\n",stud[course[i][j]].name);
       }
    }
    return 0;
}

超时代码：不超时的话下面的代码更清晰明了。

#include<stdio.h>
#include<iostream>
#include<string>
#include<algorithm>
#include<map>
#include<vector>
using namespace std;
map<int,vector<string> > Map;
bool cmp(string a,string b)
{
    return a<b;
}
int main()
{
    int i,j,n,m,k,t;
    string s;
    cin>>n>>m;
    for(i=0;i<n;i++)
    {
        cin>>s>>k;
        for(j=0;j<k;j++)
        {
            cin>>t;
            Map[t].push_back(s);
        }
    }
    for(i=1;i<=m;i++)
    {
        sort(Map[i].begin(),Map[i].end(),cmp);
        cout<<i<<" "<<Map[i].size()<<endl;
        for(j=0;j<Map[i].size();j++)
        {
            cout<<Map[i][j]<<endl;
        }
    }
    return 0;
}