PTA 1006 Sign In and Sign Out (25 分)（详细翻译）

深碍√TFBOYSˉ_ 2022-08-28 09:45 138阅读 0赞

# 背景： #

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.

**每天开始的时候，第一个在电脑室签名的人会打开门，最后一个签名离开的人会锁上门。根据签到和签出的记录，你应该找到当天开门和锁门的人。**

# 输入格式： #

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:

**每个输入文件包含一个测试用例。每个测试用例里都有一天的记录。本例以正整数M开始，M是记录的总数，单独占一行，后面是M行，每一行的格式是: ID\_number Sign\_in\_time Sign\_out\_time**

# 输出格式： #

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.

**对于每个测试用例，在一行中输出当天锁门和开锁人员的ID号。两个ID号之间必须用一个空格隔开。  
注:保证记录一致。也就是说，每个人的签到时间必须早于签到时间，并且不能有两个人同时签到或签到。**

# 输入样例： #

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

# 输出样例： #

SC3021234 CS301133

# 思路： #

本题`不涉及`复杂算法。

## C++实现代码： ##

#include <iostream>
    #include <vector>
    #include <string>
    
    using namespace std;
    
    int main()
    { 
        int M;
        cin >> M;
        vector<string> name;
        vector<int> unlock, lock;
        int m = M;
        while (M--)
        { 
            string i;
            int j = 0;
            cin >> i;
            name.push_back(i);
            cin >> i;
            j = (i[0] - '0') * 100000 + (i[1] - '0') * 10000 + (i[3] - '0') * 1000 + (i[4] - '0') * 100 + (i[6] - '0') * 10 + (i[7] - '0');
            unlock.push_back(j);
            j = 0;
            cin >> i;
            j = (i[0] - '0') * 100000 + (i[1] - '0') * 10000 + (i[3] - '0') * 1000 + (i[4] - '0') * 100 + (i[6] - '0') * 10 + (i[7] - '0');
            lock.push_back(j);
        }
        int min = 999999, max = 0, st1 = 0, st2 = 0;
        for (int i = 0; i < m; ++i)
        { 
            if (unlock[i] < min)
            { 
                min = unlock[i];
                st1 = i;
            }
            if (lock[i] > max)
            { 
                max = lock[i];
                st2 = i;
            }
        }
        cout << name[st1] << ' ';
        cout << name[st2] << endl;
        return 0;
    }