1047. Student List for Course (25)

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本博客介绍了一个课程注册统计系统的实现方案，该系统能够处理大量学生数据，并快速输出每门课程的学生名单及其数量。通过使用高效的算法和数据结构，确保了在大规模数据下也能迅速响应。

Zhejiang University has 40000 students and provides 2500 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 (<=40000), the total number of students, and K (<=2500), 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

//还是空间换时间的思路 .. 一个个匹配超时。。
//不能用string..得用char数组 ... 据说strcmp比string的compare要快。。不知道为什么..反正用string也是超时。。
#include "iostream"
#include "algorithm"
#include "vector"
#include "string"
#include "cstring"
using namespace std;
vector<char*>v[2505];
int cmp(char* a,char* b) {
	if (strcmp(a,b) < 0)
		return 1;
	return 0;
}
int main() {
	int n, k;
	int i, j;
	scanf("%d %d", &n, &k);
	for (i = 1; i <= n; i++) {
		char* name = new char[5];
		scanf("%s", name[i]);
		int c;
		scanf("%d",&c);
		for (j = 0; j < c; j++) {
			int b;
			scanf("%d", &b);;
			v[b].push_back(name);
		}
	}
	for (i = 1; i <= k; i++) {
		printf("%d %d\n", i, v[i].size());
		sort(v[i].begin(), v[i].end(), cmp);
		for (j = 0; j < v[i].size(); j++) {
			printf("%s\n", v[i][j]);
		}
	}
	for (i = 1; i <= k; i++) {
		for (j = 0; j < v[i].size(); j++) {
			delete[] v[i][j];
		}
	}
	return 0;
}