PAT 1039 (散列表的应用)

最新推荐文章于 2021-01-27 20:48:57 发布

alex1997222

最新推荐文章于 2021-01-27 20:48:57 发布

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分类专栏： ACM算法习题

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本文介绍了一种基于哈希表的学生课程注册查询系统，该系统能够高效地处理大量学生和课程的数据，通过将学生姓名转换为哈希值，实现快速查询和管理。系统接收学生查询请求，返回其注册的所有课程列表。

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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 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.

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:

解题思路:

我们可以把学生的输入姓名通过一个hash函数转换成int型作为Index存储到数组中,在调用的时候,我们可以到数组中去取得这个名

首先我们定义一个散列表

const int M = 26 * 26 * 26 * 10 + 1;  //散列表
const int N = 40010;        //总人数
vector<int> hashTable[M];  //一个散列表,姓名通过哈希表转换进行存储,课程就存储在vector<int>中

然后定义一个散列函数:

int getID(char name[]){    //散列函数
	int id = 0;
	for (int i = 0; i < 3; ++i) {
		id = id * 26 + (name[i] - 'A');
	}
	id = id * 10 + (name[3] - '0');   //转为数字的ASCII形式
	return id;
}

我们在读入学生的姓名后,将其转换成对应的ID并存储到散列表中

char name[5];    
	int n, k;    //人数,课程量
	cin >> n >> k;
	for (int i = 0; i < k; ++i) {
		int course, n;    //课程编号及人数
		cin >> course >> n;
		for (int i = 0; i < n; ++i) {
			scanf("%s", name);     //输入选课学生姓名
			int id = getID(name);
			hashTable[id].push_back(course);   /*利用哈希表对数据进行存储*/
		}
	}

之后输入对应的姓名,我们就可以到散列表中进行查询

for (int i = 0; i < n; ++i) {
		scanf("%s", name);
		int id = getID(name);   //到散列表中进行查询
		sort(hashTable[id].begin(), hashTable[id].end());  //从小到大进行排列
		printf("%s %d", name, hashTable[id].size());
		for (int j = 0; j < hashTable[id].size(); ++j) {
			printf(" %d", hashTable[id][j]);
		}
		cout << endl;
	}