问题 A: Course List for Student (25)

问题 A: Course List for Student (25)

时间限制: 1 Sec  内存限制: 32 MB
提交: 727  解决: 206
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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.

这一题利用vector可以很方便的解决问题

我们其实要做的就是把每一个课程代码存到每一个学生的选课名单里。

这样的话，学生就是一个vector，我们不断的把课程代码push进对应学生的vector中。

首先我们利用散列那一章的字符串hash，将名字这个字符串转化成唯一的整数，利用这个整数，找到hash表对应的位置。

由于是三个字符加一个字母。所以一共有26*26*26*10种名字的组合。

我们就建立这么多的vector。

const int maxn=26*26*26*10+1;//将名字唯一的对应成数字有maxn种 
vector hashTable[maxn];  //把每个名字映射在容器中，再利用容器存储课程代码 

然后不断的push课程代码到对应的hashTable中

最后查看hashTable.size()是否大于零，来进行相应的输出。

#include
#include 
#include 
#include 
using namespace std;

const int maxn=26*26*26*10+1;//将名字唯一的对应成数字有maxn种 
vector hashTable[maxn];  //把每个名字映射在容器中，再利用容器存储课程代码 

int Transfer(char s[]){  //利用算法笔记介绍的字符串hash初步提供的方法，将字符串转化成唯一的整型数字 
	int id=0;
	for(int i=0;i<3;i++){
		id=id * 26 + s[i] - 'A';
	}
	id = id * 10 + s[3] - '0';
	return id;
}

int main(){
	int n,k,i,count;
	char name[5];
	vector::iterator it;
	scanf("%d%d",&n,&k);
	for(int j=0;j