PAT--1039. Course List for Student

最新推荐文章于 2024-05-02 00:03:57 发布

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最新推荐文章于 2024-05-02 00:03:57 发布

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分类专栏： PAT 数据结构--哈希文章标签： pat

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本文介绍了一个学生课程查询系统的实现方案。针对大量学生和课程的数据处理需求，文章提出了一种利用手动哈希的方法来替代传统的map数据结构，有效解决了内存溢出的问题。通过将学生的姓名转换为26进制数并映射到哈希表中，实现了快速的学生课程注册信息查询。

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hejiang 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 (<=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.

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

题解

这题有点意思，map会爆内存不能用了呢，那就手动 hash。
把name视为26进制数，其对应的10进制数作为索引。

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

int n, k;
int a, b;

struct cmp_str{
    bool operator()(char * a, char * b){
        return strcmp(a, b) < 0;
    }
};

map<char*, set<int>, cmp_str > mp;
const int maxn = 26 * 26 * 26 * 10;
int base[] = {26 * 26 * 10, 26 * 10, 10};
set<int> h[maxn];

int Hash(char* s){
    return (s[0] - 'A') * base[0] + (s[1] - 'A') * base[1] + (s[2] - 'A') * base[2] + (s[3] - '0');
}

int main(){
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif // ONLINE_JUDGE

    scanf("%d %d", &n, &k);
    while(k--){
        scanf("%d %d", &a, &b);
        for(int i = 0; i < b; ++i){
            char* name = new char[5];
            scanf("%s", name);
//            mp[name].insert(a);
            h[Hash(name)].insert(a);
        }
    }

    for(int i = 0; i < n; ++i){
        char* name = new char[5];
        scanf("%s", name);
        int ix = Hash(name);
        printf("%s %d", name, h[ix].size());
        for(auto it = h[ix].begin(); it != h[ix].end(); ++it){
            printf(" %d", *it);
        }
       puts("");
    }
    return 0;
}