1039 Course List for Student (25分)-CFANZ编程社区

文章目录

1 题目
2 解析

2.1 题意
2.2 思路

3 参考代码

1 题目

1039 Course List for Student (25分)
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:
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

2 解析

2.1 题意

求每个学生选择的课程数目和课程编号

2.2 思路

把学生名字转换为int类型的学生编号，如果使用；map<\string,vector<\int>>将会导致超时，应该用字符串hash解决。
建立vector<\int>来存储每个学生选择课程编号，由于字符串hash的转换可知，vector<\int>最少应该有262626*10+1大小（三个字母加1个数字）；
由于数据量庞大，不要使用cin或cout；
使用二维数组存放学生编号，会导致最后一行内存超限

3 参考代码

#include 
#include 
#include 

using std::sort;
using std::vector;

const int MAXN = 26*26*26*10 + 1;//三个字母加一个数字
vector<int> student[MAXN];

int stringToNum(char name[]){
    int num = 0;
    for (int i = 0; i < 3; ++i)
    {
        num = num * 26 + (name[i] - 'A');
    }
    num = num * 10 + (name[3] - '0');
    return num;
}

int main(int argc, char const *argv[])
{
    int N, K;
    scanf("%d%d", &N, &K);

    int courseID, studentNum;
    char name[5];
    for (int i = 0; i < K; ++i)
    {
        scanf("%d%d", &courseID, &studentNum);
        for (int j = 0; j < studentNum; ++j)
        {
            scanf("%s", name);
            student[stringToNum(name)].push_back(courseID);
        }
    }

    for (int i = 0; i < N; ++i)
    {
        scanf("%s", name);
        int stundetID = stringToNum(name);
        sort(student[stundetID].begin(), student[stundetID].end());
        printf("%s %lu", name, student[stundetID].size());
        for (vector<int>::iterator it = student[stundetID].begin(); it != student[stundetID].end() ; ++it)
        {
            printf(" %d", (*it));
        }
        printf("\n");

    }
    return 0;
}