A1039 Course List for Student (25分)【C语言】

原题链接

如果使用二维数组存放学生所选课程编号，会导致最后一组数据内存超限，因此需要使用vector来减少空间消耗。

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.

输入格式：
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 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.

输出格式：
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.

输入样例：

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

输出样例：

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

实现代码：

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;const int N = 40010;
const int M = 26*26*26*10+1; //由姓名散列形成的数字上界 
vector<int> selectCourse[M]; //每个学生所选的课程编号 int getID(char name[]){int id = 0;for(int i=0; i<3; ++i){id = id * 26 + (name[i] - 'A');}id = id * 10 + (name[3] - '0');return id;
}int main()
{char name[5];int n, k; //查询人数、课程数 scanf("%d %d", &n, &k);  for(int i=0; i<k; ++i){ //对于每一门课程int course, x; //课程编号、选课人数 scanf("%d %d", &course, &x);for(int j=0; j<x; ++j){ //对于每一个选课学生 scanf("%s", name);int id = getID(name); //将姓名散列成一个整数作为编号 selectCourse[id].push_back(course); //将该课程编号加入学生散列中 }}for(int i=0; i<n; ++i){scanf("%s", name);int id = getID(name); //查询学生编号 sort(selectCourse[id].begin(), selectCourse[id].end()); //将学生所选课程，按课程号升序排列printf("%s %d", name, selectCourse[id].size());for(int j=0; j<selectCourse[id].size(); ++j){printf(" %d", selectCourse[id][j]);}printf("\n");}return 0;
}