6114 Problem A Course List for Student (25)

最新推荐文章于 2021-03-07 12:38:15 发布

fantasydreams

最新推荐文章于 2021-03-07 12:38:15 发布

阅读量524

点赞数

CC 4.0 BY-SA版权

分类专栏： C/C++ codeup OJ 文章标签： codeup 6114

本文链接：https://blog.youkuaiyun.com/fantasydreams/article/details/79279876

C/C++ 同时被 3 个专栏收录

421 篇文章

订阅专栏

389 篇文章

订阅专栏

codeup

144 篇文章

订阅专栏

本文介绍了一个基于C++实现的学生课程注册列表查询系统。该系统针对浙江大学4万名学生的2500门课程，能够快速准确地返回每位学生的选课情况。通过对学生姓名进行哈希转换并存储对应的课程编号，实现了高效的数据检索。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题 A: Course List for Student (25)
时间限制: 1 Sec 内存限制: 32 MB
献花: 97 解决: 32
[献花][花圈][TK题库]
题目描述
Zhejiang University has 40000 students and provides 2500 courses. Now given the student name lists of all the courses, you are supposed to output the registered course list for each student who comes for a query.
输入
Each input file contains one test case. For each case, the first line contains 2 positive integers: N (<=40000), the number of students who look for their course lists, and K (<=2500), the total number of courses. Then the student name lists are given for the courses (numbered from 1 to K) in the following format: for each course i, first the course index i and the number of registered students Ni (<= 200) are given in a line. Then in the next line, Ni student names are given. A student name consists of 3 capital English letters plus a one-digit number. Finally the last line contains the N names of students who come for a query. All the names and numbers in a line are separated by a space.
输出
For each test case, print your results in N lines. Each line corresponds to one student, in the following format: first print the student’s name, then the total number of registered courses of that student, and finally the indices of the courses in increasing order. The query results must be printed in the same order as input. All the data in a line must be separated by a space, with no extra space at the end of the line.
样例输入
11 5
4 7
BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
1 4
ANN0 BOB5 JAY9 LOR6
2 7
ANN0 BOB5 FRA8 JAY9 JOE4 KAT3 LOR6
3 1
BOB5
5 9
AMY7 ANN0 BOB5 DON2 FRA8 JAY9 KAT3 LOR6 ZOE1
ZOE1 ANN0 BOB5 JOE4 JAY9 FRA8 DON2 AMY7 KAT3 LOR6 NON9
样例输出
ZOE1 2 4 5
ANN0 3 1 2 5
BOB5 5 1 2 3 4 5
JOE4 1 2
JAY9 4 1 2 4 5
FRA8 3 2 4 5
DON2 2 4 5
AMY7 1 5
KAT3 3 2 4 5
LOR6 4 1 2 4 5
NON9 0

注意点：

不能使用map映射string到set集合，会超内存
字符串的格式是固定的，因此可以写一个hash规则将字符串转换为整形进行映射，时间复杂度为O(len)，len为字符串长度

#define _CRT_SECURE_NO_WARNINGS
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>

using namespace std;

const int MaxN = 180000;  // 最大('Z' - 'A') * 26 * 26 * 10 + ('Z' - 'A') * 26 * 10 + ('Z' - 'A') * 10 + '9' - '0';
vector<short> List[MaxN];

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

int main()
{
#ifdef _DEBUG
    freopen("data.txt", "r+", stdin);
#endif // _DEBUG

    int N,K,cn,id,idx;
    char name[5];
    scanf("%d %d", &N,&K);
    while (K--)
    {
        scanf("%d %d", &id, &cn);
        while (cn--)
        {
            scanf("%s", name);
            idx = StrToInt(name);
            List[idx].push_back(id);
        }
    }

    while (N--)
    {
        scanf("%s", name);
        idx = StrToInt(name);
        printf("%s %d", name, List[idx].size());
        sort(List[idx].begin(), List[idx].end());
        for (int i = 0; i < List[idx].size(); ++i)
            printf(" %d", List[idx][i]);
        if (N)
            printf("\n");
    }

    return 0;
}
/**************************************************************
    Problem: 6114
    User: Sharwen
    Language: C++
    Result: 升仙
    Time:441 ms
    Memory:11084 kb
****************************************************************/