1047 Student List for Course——PAT甲级

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:

复制代码
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:

复制代码
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

solution:

cpp 复制代码
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
int main()
{
	int n,k;cin>>n>>k;
	vector<vector<string> >v(k+1);
	for(int i=0;i<n;i++)
	{
		string name;cin>>name;
		int t;cin>>t;
		while(t--)
		{
			int cnt;cin>>cnt;
			v[cnt].push_back(name);
		}
	}
	for(int i=1;i<=k;i++)
	{
		if(v[i].size())
		{
			sort(v[i].begin(),v[i].end());
			cout<<i<<' '<<v[i].size()<<endl;
			for(int j=0;j<v[i].size();j++)
			{
				cout<<v[i][j]<<endl;
			}
		}
		else//测试点1:没人选的课程也要输出
		{
			cout<<i<<' '<<0<<endl;
		}
	}
}

一个简单的二维vector数组排序。

测试点1:没人选的课程也要输出

相关推荐
田梓燊15 分钟前
力扣:23.合并 K 个升序链表
算法·leetcode·链表
re林檎39 分钟前
算法札记——4.27
算法
AI人工智能+电脑小能手1 小时前
【大白话说Java面试题】【Java基础篇】第15题:JDK1.7中HashMap扩容为什么会发生死循环?如何解决
java·开发语言·数据结构·后端·面试·哈希算法
数据牧羊人的成长笔记1 小时前
逻辑回归与Softmax回归
算法·回归·逻辑回归
郑州光合科技余经理2 小时前
同城O2O海外版二次开发实战:从支付网关到配送算法
开发语言·前端·后端·算法·架构·uni-app·php
张健11564096483 小时前
使用信号量限制并发数量
开发语言·c++
jc06203 小时前
6.1云原生之Docker
c++·docker·云原生
Mrlxl.cn4 小时前
计算机网络——网络层
c语言·数据结构·计算机网络·考研
d111111111d5 小时前
STM32-UART封装问题解析
笔记·stm32·单片机·嵌入式硬件·学习·算法
寒秋花开曾相惜5 小时前
(学习笔记)4.2 逻辑设计和硬件控制语言HCL(4.2.1 逻辑门&4.2.2 组合电路和HCL布尔表达式)
linux·网络·数据结构·笔记·学习·fpga开发