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 (≤), the number of students who look for their course lists, and K (≤), 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​​ (≤) 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
题目分析:利用 map<string, set<int> >最后一个测试点超时了
网上说自己写一个hash可以过 超时代码
 #define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std; int main()
{
map<string, set<int> >S;
int N, K;
string s;
cin >> N >> K;
for (int i = ; i < K; i++)
{
int j, M;
cin >> j >> M;
for (int k = ; k < M; k++)
{
cin >> s;
S[s].insert(j);
}
}
for (int i = ; i < N; i++)
{
cin >> s;
cout << s << " " << S[s].size();
for (auto it : S[s])
cout << " " << it;
cout << endl;
}
}

正确代码

 #define _CRT_SECURE_NO_WARNINGS
#include <climits>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<algorithm>
#include<string>
#include<cmath>
using namespace std;
set<int> S[];
int hashF(string s)
{
int num = ;
for (int i = ; i < ; i++)
num = num * + s[i] - 'A';
num = num * + s[] - '';
return num;
}
int main()
{
int N, K;
string s;
cin >> N >> K;
for (int i = ; i < K; i++)
{
int j, M;
cin >> j >> M;
for (int k = ; k < M; k++)
{
cin >> s;
S[hashF(s)].insert(j);
}
}
for (int i = ; i < N; i++)
{
cin >> s;
cout << s << " " << S[hashF(s)].size();
for (auto it : S[hashF(s)])
cout << " " << it;
cout << endl;
}
}

1039 Course List for Student (25分)的更多相关文章

  1. PAT 甲级 1039 Course List for Student (25 分)(字符串哈希,优先队列,没想到是哈希)*

    1039 Course List for Student (25 分)   Zhejiang University has 40000 students and provides 2500 cours ...

  2. PAT 1039 Course List for Student (25分) 使用map<string, vector<int>>

    题目 Zhejiang University has 40000 students and provides 2500 courses. Now given the student name list ...

  3. 【PAT甲级】1039 Course List for Student (25 分)(vector嵌套于map,段错误原因未知)

    题意: 输入两个正整数N和K(N<=40000,K<=2500),分别为学生和课程的数量.接下来输入K门课的信息,先输入每门课的ID再输入有多少学生选了这门课,接下来输入学生们的ID.最后 ...

  4. 1039. Course List for Student (25)

    题目链接:http://www.patest.cn/contests/pat-a-practise/1039 题目: 1039. Course List for Student (25) 时间限制 2 ...

  5. PAT甲题题解-1039. Course List for Student (25)-建立映射+vector

    博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6789157.html特别不喜欢那些随便转载别人的原创文章又不给 ...

  6. PAT (Advanced Level) 1039. Course List for Student (25)

    map会超时,二分吧... #include<iostream> #include<cstring> #include<cmath> #include<alg ...

  7. A1039 Course List for Student (25 分)

    一.技术总结 这里由于复杂度的限制,只能够使用vector,然后进行字符串转化:考虑到string.cin.cout会超时,可以使⽤用hash(262626*10+10)将学⽣生姓名变为int型,然后 ...

  8. PAT 甲级 1047 Student List for Course (25 分)(cout超时,string scanf printf注意点,字符串哈希反哈希)

    1047 Student List for Course (25 分)   Zhejiang University has 40,000 students and provides 2,500 cou ...

  9. PAT 1039 Course List for Student[难]

    1039 Course List for Student (25 分) Zhejiang University has 40000 students and provides 2500 courses ...

随机推荐

  1. Hadoop fs 基础命令

    操作hdfs的基本命令 在hdfs中,路径需要用绝对路径 1. 查看根目录 hadoop fs -ls / 2. 递归查看所有文件和文件夹 -lsr等同于-ls -R hadoop fs -lsr / ...

  2. JetBrains全系列产品2019.3.2注解教程

    1.JetBrains官方网站 https://www.jetbrains.com/ JetBrains是一家捷克的软件开发公司 IDE工具: * IntelliJ IDEA    一套智慧型的Jav ...

  3. php实现post跳转

    大家否知道php可以利用header('Location')实现get请求跳转. php利用curl可以实现模拟post请求. 但是却找不到php现成的实现post跳转.那么问题来了,如果有这个需求该 ...

  4. (转)协议森林07 傀儡 (UDP协议)

    协议森林07 傀儡 (UDP协议) 作者:Vamei 出处:http://www.cnblogs.com/vamei 欢迎转载,也请保留这段声明.谢谢! 我们已经讲解了物理层.连接层和网络层.最开始的 ...

  5. JavaScript 模式》读书笔记(3)— 字面量和构造函数2

    上一篇啊,我们聊了聊字面量对象和自定义构造函数.这一篇,我们继续,来聊聊new和数组字面量. 三.强制使用new的模式 要知道,构造函数,只是一个普通的函数,只不过它却是以new的方式调用.如果在调用 ...

  6. ADO.NET 的使用(一)

    一.ADO.NET概要 ADO.NET 是一组向 .NET Framework 程序员公开数据访问服务的类. ADO.NET 为创建分布式数据共享应用程序提供了一组丰富的组件. 它提供了对关系数据.X ...

  7. Journal of Proteome Research | 人类牙槽骨蛋白的蛋白质组学和n端分析:改进的蛋白质提取方法和LysargiNase消化策略增加了蛋白质组的覆盖率和缺失蛋白的识别 | (解读人:卜繁宇)

    文献名:Proteomic and N-Terminomic TAILS Analyses of Human Alveolar Bone Proteins: Improved Protein Extr ...

  8. thinkphp 前后端分离

    thinkphp 前后端分离 简单记录一下之前学习tp的历程吧. 前端HTML页面渲染 <?php namespace app\index\controller; use think\Contr ...

  9. python 顺序读取文件夹下面的文件(自定义排序方式)

    我们在读取文件夹下面的文件时,有时是希望能够按照相应的顺序来读取,但是 file_lists=os.listdir()返回的文件名不一定是顺序的,也就是说结果是不固定的.就比如读取下面这些文件,希望能 ...

  10. python基础知识1——简介与入门

    什么是Python:Python能做什么:安装与更新:第一个Python程序:变量:pyc字节码:编码:条件和循环:Python运算符:算数,比较,赋值,位,逻辑::::::::::::::::::: ...