1006 Sign In and Sign Out （25 分）

最新推荐文章于 2022-08-12 23:25:22 发布

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本文介绍了一种算法，用于从给定的签到记录中找出每天第一个进入和最后一个离开计算机房的人。通过将时间转换为秒进行比较，该算法能够快速准确地找到目标人员。

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

解题技巧：这个题无非就是找出谁最早到房间，谁最晚离开房间，将所有人的时间换算成秒比较即可。

代码如下：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int M, Start = 24*3600, End = 0;
	string sta, end;
	cin >> M;
	for (int i = 0; i < M; i++)
	{
		string ID;
		cin >> ID;

		int HS, MS, SS, HE, ME, SE;
		scanf("%d:%d:%d %d:%d:%d", &HS, &MS, &SS, &HE, &ME, &SE);

		int st = HS * 3600 + MS * 60 + SS, et = HE * 3600 + ME * 60 + SE;
		if (st < Start)
		{
			Start = st;
			sta = ID;
		}
		if (et > End)
		{
			End = et;
			end = ID;
		}
	}
	cout << sta << " " << end;
	return 0;
}