PTA(Advanced Level)1047.Student List for Course

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

思路

• 在读取的时候，无法确定一个课程会有多少学生选，比较稳妥的方案是开vector<int>存储选秀对应课程的学生的信息
• 信息有2种选择，一种是存储学生本身的名字，但是因为本题还要排序，会比较耗时。另一种是另开char names[][6]数组，在vector里存储下标，要输出的时候再去对应找

代码

#include<bits/stdc++.h>
using namespace std;
vector<int> courses[2510];
char names[40010][6];
bool cmp(int a, int b)
{
	return strcmp(names[a], names[b]) < 0;
}	//因为vector存储的是下标，所以是根据下标来对字符串进行排序
int main()
{
	int n, k;
	scanf("%d%d", &n, &k);

	int cnt, tmp;
	for(int i=0;i<n;i++)
	{
		scanf("%s", &names[i]);
		scanf("%d", &cnt);
		for(int j=0;j<cnt;j++)
		{
			scanf("%d", &tmp);
			courses[tmp].push_back(i);	//在相应的课程里存储对应的学生的ID(根据ID去names里可以查询到名字)
		}
	}
	for(int i=1;i<=k;i++)
	{
		sort(courses[i].begin(), courses[i].end(), cmp);
		printf("%d %d
", i, courses[i].size());
		for(int j=0;j<courses[i].size();j++)
            printf("%s
", names[courses[i][j]]);
	}
	return 0;
}

引用

https://pintia.cn/problem-sets/994805342720868352/problems/994805433955368960