A1006 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

用devc++时候用#include<stdio.h>,c编译通过，但是在PAT上编译错误，改成c++后编译通过。。。求解惑

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
struct pNode{
	char id[20];
	int hh,mm,ss;	
}node,first,last;
int cmp(pNode temp1,pNode temp2){
	if(temp1.hh>temp2.hh) return 1;     //temp1比temp2晚 
	else if(temp1.hh<temp2.hh) return 2; //temp1比temp2早 
	else{
		if(temp1.mm>temp2.mm) return 1;
	    else if(temp1.mm<temp2.mm) return 2;
	    else{
	    		if(temp1.ss>temp2.ss) return 1;
	            else  return 2;
		}
	}
	 
}
int main(){
	int n;
	last.hh=0,first.mm=0,first.ss=0;
	first.hh=23,last.mm=60,last.ss=60;
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%s ",node.id);
		scanf("%d:%d:%d ",&node.hh,&node.mm,&node.ss);
		if(cmp(node,first)==2) first=node;
		scanf("%d:%d:%d",&node.hh,&node.mm,&node.ss);
		if(cmp(node,last)==1) last=node;
	}
	printf("%s %s",first.id,last.id);
	return 0;
}

另：可以将日期转成int型直接进行比较。。

来源：CSDN

作者：后入赵丽颖

链接：https://blog.csdn.net/qq_36491095/article/details/104315199