1047 Student List for Course (25 分)

本篇博客介绍了一个算法挑战,即统计浙江大学4万名学生在2500门课程中的注册情况。通过输入每个学生的注册课程列表,算法需输出每门课程的学生名单。详细介绍了输入输出规格,提供了一个C++代码示例,展示了如何使用哈希映射和排序来解决这个问题。

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

C++:

 

/*
 @Date    : 2018-01-27 20:12:16
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1047
 */

#include <iostream>
#include <cstdio>
#include <string>
#include <vector>
#include <algorithm>

using namespace std;

int name2num(char name[])
{
    return (name[0] - 'A') * 26 * 26 * 10 + (name[1] - 'A') * 26 * 10 +
    (name[2] - 'A') * 10 + (name[3] - '0');
}

void num2name(char name[], int num)
{
    name[3] = num % 10 + '0';
    num /= 10;
    name[2] = num % 26 + 'A';
    num /= 26;
    name[1] = num % 26 + 'A';
    num /= 26;
    name[0] = num % 26 + 'A';
    return;
}

struct Course
{
    vector<int> students;
};

int main(int argc, char const *argv[])
{
    int N, K;
    scanf("%d %d", &N, &K);
    char name[5];
    int C, c;
    Course courseData[2501];
    for (int i = 0; i != N; ++i)
    {
        scanf("%s %d", name, &C);
        for (int j = 0; j != C; ++j)
        {
            scanf("%d", &c);
            courseData[c].students.push_back(name2num(name));
        }
    }
    for (int i = 1; i <= K; ++i)
    {
        printf("%d %d\n", i, courseData[i].students.size());
        sort(courseData[i].students.begin(), courseData[i].students.end());
        for (const auto id : courseData[i].students)
        {
            num2name(name, id);
            printf("%s\n", name);
        }
    }
    return 0;
}

 

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