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题目链接1006 Sign In and Sign Out (25 分)

题干大意

根据时间，输出最早来和最晚走的人的id

思路

输入时即用earliest和latest处理保存当前已经输入的人中的最早和最晚

AC代码

C++
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#include <iostream>
using namespace std;

struct employee {
    string id;
    int    hour{}, minute{}, second{};

} temp, earliest, latest;

bool great(employee& a, employee& b)
{
    if (a.hour != b.hour)
        return a.hour > b.hour;
    else if (a.minute != b.minute)
        return a.minute > b.minute;
    else
        return a.second > b.second;
}

int main()
{
    int n;
    cin >> n;
    earliest.hour = 24, earliest.minute = 60, earliest.second = 60;
    latest.hour = latest.minute = latest.second = 0;
    while (n--) {
        char c;
        cin >> temp.id >> temp.hour >> c >> temp.minute >> c >> temp.second;
        if (!great(temp, earliest))
            earliest = temp;
        cin >> temp.hour >> c >> temp.minute >> c >> temp.second;
        if (great(temp, latest))
            swap(temp, latest); // 注意用swap的话，只能交换一次，因为有两个值需要更新
    }
    cout << earliest.id << " " << latest.id;
    return 0;
}