1039 Course List for Student

最新推荐文章于 2024-12-02 11:52:39 发布

Win_Man

最新推荐文章于 2024-12-02 11:52:39 发布

阅读量338

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT (Advanced Level) 文章标签： pat

本文链接：https://blog.youkuaiyun.com/Win_Man/article/details/50640308

PAT (Advanced Level) 专栏收录该内容

68 篇文章

订阅专栏

本文介绍了一个学生选课查询系统的实现方法，采用哈希表技术优化查询速度，通过将学生姓名转换为哈希值来快速查找每位学生的选课记录，并提供了一段C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

Input Specification:

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 N_i (<= 200) are given in a line. Then in the next line, N_i 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.

Output Specification:

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.

Sample Input:

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

Sample Output:

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
解题思路：一开始以为是一题挺简单的题目，用一下c++中的map很轻松就能过吧，写了一下，发现最后一个case不能过，各种简单的优化都上了，还是没有用，最后网上看了下别人的解题思路，发现说，用map进行查询的时候检索名字比较慢，题目中给定名字的组成是三个大写字母加一个数字，总共26*26*26*10中可能性，用哈希来表示一下名字的话，索引的时候查询效率就变成O（1）了
#include<iostream>
#include<map>
#include<vector>
#include<string>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main(){
	for (int query_num, course_num; scanf("%d%d",&query_num,&course_num)!=EOF;){
		//构建名字的哈希表
		vector<vector<int>>names(26*26*26*10 + 5);
		while (course_num--){
			int course, stu_num;
			//cin >> course >> stu_num;
			scanf("%d%d", &course, &stu_num);
			char name[5];
			while (stu_num--){
				scanf("%s", name);
				int index = (name[0] - 'A') * 26 * 26 * 10 + (name[1] - 'A') * 26 * 10 + (name[2] - 'A') * 10 + (name[3] - '0');
				names[index].push_back(course);
			}
		}
		while (query_num--){
			char query_name[5];
			scanf("%s", query_name);
			int index = (query_name[0] - 'A') * 26 * 26 * 10 + (query_name[1] - 'A') * 26 * 10 + (query_name[2] - 'A') * 10 + (query_name[3] - '0');
			sort(names[index].begin(),names[index].end());
			printf("%s %d", query_name, names[index].size());
			for (int i = 0; i < names[index].size(); i++){
				printf(" %d", names[index][i]);
			}
			printf("\n");
		};
	}
	return 0;
}