Hdu1234 door opener and door closer (water question)

Topic link     http://acm.hdu.edu.cn/showproblem.php?pid=1234

Question

Problem Description

The first person to the computer room every day should open the door , The last person to leave is to close the door . There are a lot of messy machine room signs
To 、 Check out record , Please find out who opened and closed the door according to the record .

Input

The first line of the test input gives the total number of days recorded N ( > 0 ). The following is listed N A record of days .
The daily record gives the number of entries in the first line M ( > 0 ), Here is M That's ok , The format of each line is :
ID number Check in time Check out time
The time is according to “ Hours : minute : Second ”（ Each account 2 position ） give , The ID number is no longer than 15 String .

Output

Output the record of each day 1 That's ok , That is, the ID number of the person who opened and closed the door that day , Intermediate use 1 The blank space to separate .
Be careful ：

In the referee's standard test input , All records shall be complete , Everyone's check-in time is before the check-out time ,
And there is no situation that many people sign in or sign out at the same time .

Solve

Find the smallest and largest in a pile of time , Just pay attention to format control

AC Code 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int t, n;
int main(void)
{
	scanf("%d", &t);
	while(t--){
		scanf("%d", &n);
		string en, st;	            // Starting and ending Certificate No  
		int t_en = -1, t_st = 1e9;	// Starting and ending time  
		int h1, h2, m1, m2, s1, s2;
		for(int i = 1; i <= n; i++){
			string s; cin >>s;
			// Read in time  
			scanf("%02d:%02d:%02d", &h1, &m1, &s1);
			scanf("%02d:%02d:%02d", &h2, &m2, &s2);
			// Calculation  
			int t1 = h1 * 3600 + m1 * 60 + s1;
			int t2 = h2 * 3600 + m2 * 60 + s2;
			// Compare  
			if(t1 < t_st) { st = s; t_st = t1; }
			if(t2 > t_en) { en = s; t_en = t2; } 
		}
		cout <<st <<" " <<en <<endl;
	}
	return 0;
}