A1039. Course List for Student-优快云博客

本文探讨了一种高效的学生选课系统实现方法，针对大规模学生和课程数据，提出了使用哈希函数映射代替传统二维数组存储策略，有效解决了数据处理过程中的超时问题。通过自定义哈希函数，实现了学生姓名到数组下标的快速映射，避免了频繁的字符串转换和map查询操作，显著提升了系统性能。

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 (<=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 N_i (<= 200) 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

总结：

1、由于学生数太多，再乘以课程数之后数量过大，因此不能使用二维数组存储选课信息（一个学生一行）。方案一：以学生为主体记录选课信息，使用map<string, set<int> >进行以学生名字为键的存储方法。但导致了最后一组数据超时。

2、方案一要同时使用map，string，加之数据过大导致超时。因此方案二可以直接自己写一个hash函数，映射 char[ ] 到stu数组下标的关系。这样省去了char [ ] 与string之间的转换，也不需要再使用map进行查询操作。

3、map[ " key" ] 在查询过程中必须先确定是否存在 if(mp.count( "key" ) == 0)，否则会返回错误的结果（没有该键值时，会在map中插入该key，并将其值设为默认值）。

4、字符串映射为int的方法：若只有小写字母，则将其当成26进制数，计算出int。若字符串为字母加数字，则映射成36进制。如果数字仅仅在固定位置（如本题），则先将字母计算为26进制int数，再将其 * 10 + 末位数字。

超时代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<string>
 5 #include<vector>
 6 #include<map>
 7 using namespace std;
 8 string search_[400000];
 9 vector<int> stu[400000];
10 map<string, int> mp;
11 bool cmp(int a, int b){
12     return a < b;
13 }
14 int main(){
15     int N, K, cIndex, Ni, stuId = 0;
16     string temp;
17     char id[5];
18     scanf("%d%d", &N, &K);
19     for(int i = 0; i < K; i++){
20         scanf("%d %d", &cIndex, &Ni);
21         for(int j = 0; j < Ni; j++){
22             int pt;
23             scanf("%s", id);
24             temp = id;
25             if(mp.count(temp) == 0){
26                 pt = stuId;
27                 mp[temp] = stuId++;
28             }else{
29                 pt = mp[temp];
30             }
31             stu[pt].push_back(cIndex);
32         }
33     }
34     for(int i = 0; i < N; i++){
35         scanf("%s", id);
36         temp = id;
37         search_[i] = temp;
38     }
39     for(int i = 0; i < N; i++){
40         printf("%s", search_[i].c_str());
41         if(mp.count(search_[i]) == 0){
42             printf(" 0\n");
43             continue;
44         }
45         int pt = mp[search_[i]];
46         sort(stu[pt].begin(), stu[pt].end());
47         int len = stu[pt].size();
48         printf(" %d", len);
49         for(int j = 0; j < len; j++)
50             printf(" %d", stu[pt][j]);
51         printf("\n");
52     }
53     return 0;
54 }

View Code

正确代码：

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<vector>
 5 #include<string.h>
 6 using namespace std;
 7 vector<int> stu[26 * 26 * 26 * 10 + 1];
 8 bool cmp(int a, int b){
 9     return a < b;
10 }
11 long long hash_(char str[]){
12     int len = strlen(str);
13     long long ans = 0, P = 1;
14     for(int i = len - 2; i >= 0; i--){
15         ans += (str[i] - 'A') * P;
16         P *= 26;
17     }
18     ans = ans * 10 + str[len - 1] - '0';
19     return ans;
20 }
21 int main(){
22     int N, K, cIndex, Ni, stuId = 0;
23     string temp;
24     char id[5];
25     scanf("%d%d", &N, &K);
26     for(int i = 0; i < K; i++){
27         scanf("%d %d", &cIndex, &Ni);
28         for(int j = 0; j < Ni; j++){
29             long long pt;
30             scanf("%s", id);
31             pt = hash_(id);
32             stu[pt].push_back(cIndex);
33         }
34     }
35     for(int i = 0; i < N; i++){
36         scanf("%s", id);
37         printf("%s", id);
38         long long pt = hash_(id);
39         sort(stu[pt].begin(), stu[pt].end());
40         int len = stu[pt].size();
41         printf(" %d", len);
42         for(int j = 0; j < len; j++)
43             printf(" %d", stu[pt][j]);
44         printf("\n");
45     }
46     return 0;
47 }