题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805516654460928
题面：
1006 Sign In and Sign Out (25 分)
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
水题，利用了string方便的比较功能，没有遇到什么坑/
大致题意：
给你M条记录，每条记录包含了记录id、签到时间、签退时间这三个数据，然后让你输出最早签到和最晚签退（即最早进来、最晚出去）的两个人的id。
解题思路：
利用string方便的字符串比较机制，例如 string a="10:00:00",b="24:59:59"; 那么a<b。【原理请自行百度】
声明两个string变量 start和end分别存储最早进来和最晚出去的时间，对其赋初值，令start="24:59:59",end="00:00:00"。
然后m次循环，每次循环判断当前签到时间是否小于start，当前签退时间是否大于end。如果是的话保存相应的id。代码如下
#include <bits/stdc++.h>
using namespace std;
int main(){
string in,out,s,e,id;
string start="24:59:59",end="00:00:00";
int m;
cin>>m;
while(m--){
cin>>id>>s>>e;
if(s<=start){
start=s;
in=id;
}
if(e>=end){
end=e;
out=id;
}
}
cout<<in<<" "<<out;
return 0;
}