题目例如以下：
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
这是一道比較简单的统计、输出题目，由于输出不涉及查询，而是顺序输出。因此不必使用map进行倒排索引。仅仅须要给每门课一个vector<char*>，把全部选这门课的人压入对应容器，然后按字典序排序，最后输出。
须要注意的是。对于字符处理的题目，要尽量使用char*而避免使用string，尽管string非常方便，可是对于string的比較和拷贝操作会消耗大量时间。easy超时。
char*没有string好用，可是在c++特性下我么能够使用char* name = new char[4];来创建一个刚好容纳名字的空间，然后把它压入容器中。
对于char*的比較，刚開始我使用的是strcmp，可是它是s1<s2返回-1，s1==s2返回0，s1>s2返回1，而sort函数默认使用<比較，返回1代表满足，因此二者不是非常匹配，我最后採用了手动比較每一位的方法。
代码例如以下：
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <map>
#include <vector>
#include <string>
#include <string.h>
#include <algorithm>
using namespace std;
int compare(const char *name1, const char *name2){
for(int i = 0; i < 4; i++){
if(name1[i] < name2[i]){
return 1;
}else if(name1[i] == name2[i]){
continue;
}else{
return 0;
}
}
}
int main()
{
int N,K,C,num;
cin >> N >> K;
vector<vector<char*> > course(K+1);
for(int i = 0; i < N; i++){
char* name = new char[4];
scanf("%s%d",name,&C);
for(int j = 0; j < C; j++){
scanf("%d",&num);
course[num].push_back(name);
}
}
for(int j = 1; j <= K; j++){
sort(course[j].begin(),course[j].end(),compare);
printf("%d %d\n",j,course[j].size());
for(int k = 0; k < course[j].size(); k++){
printf("%s\n",course[j][k]);
}
}
return 0;
}