1039 Course List for Student (25分) C++ 优化

最新推荐文章于 2024-12-02 11:52:39 发布

原创最新推荐文章于 2024-12-02 11:52:39 发布 · 100 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

PTA刷题专栏收录该内容

90 篇文章

订阅专栏

本文介绍了一个学生课程注册查询系统的设计与实现，该系统能够处理大量学生和课程数据，通过优化输入输出和数据结构，实现了高效的学生课程列表查询功能。

1039 Course List for Student (25分)

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 (≤40,000), the number of students who look for their course lists, and K (≤2,500), 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(≤200) are given in a line. Then in the next line, Nistudent 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

如果用set，同时cin，cout，scanf，printf并用，可能会超时，也可能不超时：
在这里插入图片描述

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
map<string,set<int>> stu;
int n,k;
int main()
{
	string name;
	cin>>n>>k;
	for(int i=0;i<k;i++)
	{
		int num,total;
		scanf("%d%d",&num,&total);
		for(int j=0;j<total;j++)
		{
			cin>>name;
			stu[name].insert(num);	
		}		
	}
	for(int i=0;i<n;i++)
	{
		cin>>name;
		cout<<name;
		printf(" %d",stu[name].size());
		for(auto x:stu[name]) printf(" %d",x);
		printf("\n");
	}
	return 0;
}

把set变为vector后就没有问题了。同时加入输入输出优化语句：ios::sync_with_stdio(false); cin.tie(0);
在这里插入图片描述

#include<bits/stdc++.h>
using namespace std;

unordered_map<string, vector<int>> students;

int main()
{
    ios::sync_with_stdio(false); //语句ios::sync_with_stdio(false);这样就可以取消cin于stdin的同步，大大加快运行速度了。但是此时不能使用scanf和printf语句，否则会出现问题。
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    while (k -- )
    {
        int id, m;
        cin >> id >> m;
        while (m -- )
        {
            string name;
            cin >> name;
            students[name].push_back(id);
        }
    }
    while (n -- )
    {
        string name;
        cin >> name;
        auto& ls = students[name];
        cout << name << ' ' << ls.size();
        sort(ls.begin(), ls.end());
        for (auto l : ls) cout << ' ' << l;
        cout << endl;
    }
    return 0;
}