分数 25
作者 CHEN, Yue
单位 浙江大学

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

代码长度限制
16 KB
时间限制
400 ms
内存限制
64 MB

题目翻译

背景：
每天开始的时候，第一个在电脑室签名的人会打开门，最后一个签名离开的人会锁上门。根据签到和签出的记录，你应该找到当天开门和锁门的人。
输入格式：
每个输入文件包含一个测试用例。每个测试用例里都有一天的记录。本例以正整数M开始，M是记录的总数，单独占一行，后面是M行，每一行的格式是: ID_number Sign_in_time Sign_out_time
输出格式：
对于每个测试用例，在一行中输出当天锁门和开锁人员的ID号。两个ID号之间必须用一个空格隔开。
注:保证记录一致。也就是说，每个人的签到时间必须早于签到时间，并且不能有两个人同时签到或签到。

代码

#include <iostream>
#include <string>using namespace std;int main(){int m;string unlock="99:99:99",lock="00:00:00",unlockID,lockID,tempID,tempStart,tempEnd;cin>>m;while (m--) {cin>>tempID>>tempStart>>tempEnd;if (tempStart<unlock){unlock=tempStart;unlockID=tempID;}if(tempEnd>lock){lock=tempEnd;lockID=tempID;}}cout << unlockID << " " << lockID;
}