pat

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.

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.

3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40

SC3021234 CS301133

题目大意：每天来的最早的人开门，走的最晚的人锁门。给出每个人的ID和出入时间，求出开门的和锁门的人的ID。

分析：直接将时间当成字符串直接进行比较即可。

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using namespace std;
int (){
    int n;
    string id,in,out,first="24:00:00",last="00:00:00",UnlockId,LockId;
    cin >> n;
    for (int i=0;i<n;i++){
        cin >> id >> in >> out;
        if (in < first){
            UnlockId = id;
            first = in;
        }
        if (out > last){
            LockId = id;
            last = out;
        }
    }
    cout << UnlockId << " " << LockId;
    return 0;
}