PAT1006题解

PAT1006题解

At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day.

Input Specification:

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time
where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters.

Output Specification:

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input:
3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40
Sample Output:
SC3021234 CS301133

题目问什么：
简单说就是，第一个来的学生开门，最后一个走的学生锁门。现在知道了学生的ID以及他们来和走的时间，请写算法计算谁开门，谁锁门。

分析：首先根据题意，我们可以构造一个结构体来把学生表示出来。主要就是三个信息：

• ID
• 进房间的时间
• 离开房间的时间

再考虑用vector来存储这些结构体，因为这里需要知道时间的先后，那么必然牵涉到排序。

所以，笔者认为，这题如果用C++ STL解，考察的知识点就是对sort函数的应用。

#include <iostream>#include <string>#include <vector>#include <algorithm> // 在oj上需要引入这个头文件using namespace std;typedef struct Node{    string ID;    string signInTime;    string signOutTime;} Student;// 只需要写两个比较函数即可bool compare_in_time(Student a, Student b){    return a.signInTime < b.signInTime; //升序排列}bool compare_out_time(Student a, Student b){    return a.signOutTime > b.signOutTime; //降序排列}int main(){    int M;    cin >> M;    vector<Student> stus;    for(int i = 0; i < M; i++)    {        Student student; // 存储在栈空间（相对于堆）中        cin >> student.ID >> student.signInTime >> student.signOutTime;        stus.push_back(student);    }    sort(stus.begin(), stus.end(),compare_in_time);    cout << stus[0].ID << " ";    sort(stus.begin(), stus.end(),compare_out_time);    cout << stus[0].ID << endl;    return 0;}

对string排序，是按照字典序来，所以直接比较即可。