PTA甲级——1006

1006 Sign In and Sign Out

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:

1
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:

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

Sample Output:

1
SC3021234 CS301133

思路

​ 大致题意：给你一些时间，然后让你对这些时间进行排序，从早到晚，然后输出前两名的id编号。

​ 思路：基本也是一些非常简单的思路，可以选择使用一个结构体存一个时间的id和时间的两个部分然后再自定义排排序就ok了。

代码

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#include <iostream>
#include <algorithm>

using namespace std;
const int N = 1e5 + 10;

typedef struct {
    string s1;
    string s2;
    string s3;
}Node;
Node node[N];
int n, k;

int main()
{
    ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
        string s1, s2, s3;
        cin >> s1 >> s2 >> s3;
        node[k].s1 = s1;
        node[k].s2 = s2;
        node[k++].s3 = s3;
    }
    
    sort(node, node + k, [](Node a, Node b){return a.s2 < b.s2;});   
    cout << node[0].s1 <<" ";

    sort(node, node + k, [](Node a, Node b){return a.s3 > b.s3;});
    cout << node[0].s1;

    cout << '\n';
    return 0;
}