1047 Student List for Course (25分)【vector】

1047 Student List for Course (25分)

Zhejiang University has 40,000 students and provides 2,500 courses. Now given the registered course list of each student, you are supposed to output the student name lists of all the courses.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 numbers: N (≤40,000), the total number of students, and K (≤2,500), the total number of courses. Then N lines follow, each contains a student's name (3 capital English letters plus a one-digit number), a positive number C (≤20) which is the number of courses that this student has registered, and then followed by C course numbers. For the sake of simplicity, the courses are numbered from 1 to K.

Output Specification:

For each test case, print the student name lists of all the courses in increasing order of the course numbers. For each course, first print in one line the course number and the number of registered students, separated by a space. Then output the students' names in alphabetical order. Each name occupies a line.

Sample Input:

10 5
ZOE1 2 4 5
ANN0 3 5 2 1
BOB5 5 3 4 2 1 5
JOE4 1 2
JAY9 4 1 2 5 4
FRA8 3 4 2 5
DON2 2 4 5
AMY7 1 5
KAT3 3 5 4 2
LOR6 4 2 4 1 5

Sample Output:

1 4
ANN0
BOB5
JAY9
LOR6
2 7
ANN0
BOB5
FRA8
JAY9
JOE4
KAT3
LOR6
3 1
BOB5
4 7
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1
5 9
AMY7
ANN0
BOB5
DON2
FRA8
JAY9
KAT3
LOR6
ZOE1

解题思路：

给出选课人数和课程数目，然后再给出每个人的选课情况，请针对每门课程输出选课人数以及所有选该课的学生姓名。

我们首先利用一个stu[maxn][5]数组来存放每个学生的姓名，其中stu[i]表示第i个学生。然后再利用一个vector数组courses来存放选每门课的学生编号，其中courses[i]表示第i门课的选课学生。

注意：这里最好不要使用string，像这样数据范围很大的情况，使用string一般容易超时。

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
#define maxn 40010
#define maxk 2510

char stu[maxn][5];                      //每个学生姓名
vector<int> courses[maxk];       //第i门课的所有学生编号

int cmp(int a, int b)
{
	return strcmp(stu[a], stu[b]) < 0;    //按姓名的字典顺序排序
}

int main()
{
	int n, k;
	scanf("%d %d", &n, &k);
	for (int i = 0; i < n; i++)
	{
		int c;
		scanf("%s %d", stu[i], &c);
		for (int j = 0; j < c; j++)
		{
			int course;
			scanf("%d", &course);
			courses[course].push_back(i);
		}
	}
	for (int i = 1; i <= k; i++)
	{
		printf("%d %d\n", i, courses[i].size());
		sort(courses[i].begin(), courses[i].end(), cmp);     //对第i门课的学生排序
		for (int j = 0; j < courses[i].size(); j++)
			printf("%s\n", stu[courses[i][j]]);
	}
	return 0;
}