PAT-A1006 Sign In and Sign Out (25分)

原题

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

分类

循环、简单数值处理

题意说明

第一个到机房的人要开门，最后一个签退的人要锁门。给定签到和签退记录，找出当天开门和锁门的人。

思路分析

这道题似乎过于简单，不像是25分的题。只要遍历出勤记录，找到最早到达时间和最晚离开时间，并记录对应人员id就可以了。

参考代码

#include <cstdio>
#include <iostream>
#include <string>

using namespace std;
const int INF = 1000000000;

int convertTime(int h, int m, int s) {
    return h * 60 * 60 + m * 60 + s;
}

int main() {
    //共n条记录
    int n, h, m, s;
    scanf("%d", &n);
    int minIn = INF, maxOut = -1, inTime, outTime;
    string id, minId, maxId;
    for (int i = 0; i < n; ++i) {
        cin >> id;
        scanf("%d:%d:%d", &h, &m, &s);
        inTime = convertTime(h, m, s);
        if (minIn > inTime) {
            minIn = inTime;
            minId = id;
        }
        scanf("%d:%d:%d", &h, &m, &s);
        outTime = convertTime(h, m, s);
        if (maxOut < outTime) {
            maxOut = outTime;
            maxId = id;
        }
    }

    cout << minId << " " << maxId << endl;
    return 0;
}

来源：CSDN

作者：iff2080

链接：https://blog.csdn.net/qq_36382924/article/details/104068765

标签

sign