1006 Sign In and Sign Out (25 point(s))-优快云博客

1006 Sign In and Sign Out (25 point(s))

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

注意点：

1. 表示时间的字符串按照固定格式输入，所以可以直接按照string字符串规则比较；

2. 这道题这样做麻烦了，复杂度 $o(nlogn)$ ，用于按照不同cmp规则排序，实际上在输入的时候比较即可，复杂度 $o(n)$ 。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct Record{
	string id_num;
	string in;
	string out;
};
bool cmp1(Record a,Record b){
	return a.in<b.in;
}
bool cmp2(Record a,Record b){
	return a.out<b.out;
}
int main(void){
	int n;
	Record r[1010];
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>r[i].id_num>>r[i].in>>r[i].out;
	}
	sort(r,r+n,cmp1);
	cout<<r[0].id_num<<" ";
	sort(r,r+n,cmp2);
	cout<<r[n-1].id_num<<endl;
	return 0;
}