題目例如以下：

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

這是一道比較簡單的統計、輸出題目，由于輸出不涉及查詢，而是順序輸出。因此不必使用map進行倒排索引。僅僅須要給每門課一個vector<char*>，把全部選這門課的人壓入對應容器，然后按字典序排序，最后輸出。

須要注意的是。對于字符處理的題目，要盡量使用char*而避免使用string，盡管string非常方便，可是對于string的比較和拷貝操作會消耗大量時間。easy超時。

char*沒有string好用，可是在c++特性下我么能夠使用char* name = new char[4];來創建一個剛好容納名字的空間，然后把它壓入容器中。

對于char*的比較，剛開始我使用的是strcmp，可是它是s1<s2返回-1，s1==s2返回0，s1>s2返回1，而sort函數默認使用<比較，返回1代表滿足，因此二者不是非常匹配，我最后採用了手動比較每一位的方法。

代碼例如以下：

#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; }

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