[PAT (Advanced Level) ]1006. Sign In and Sign Out解题文档

1006. Sign In and Sign Out (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

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

Sample Output:

SC3021234 CS301133

分析：

其实此题就是比大小，可能有些PATer会采用先比较“时”，再比较“分”，最后比较“秒”的策略，笔者并不认为这样好。完全可以把“：”变为0，比如08：00：00这个字符串转为“08000000”,再转化为int或者long，就变成了“8000000”,此时就是比大小了，不需要太多的麻烦。

贴上C++代码如下。

#include<iostream>#include<string>#include<strstream>#include<vector>using namespace std;int N;long transtoInt(string str){    for(int i=0;i<str.size();i++){        if(str[i]==':'){            str[i]='0';        }    }    return atol(str.c_str());}int main(){    cin>>N;    vector<long> vs;//start    vector<long> ve;//end    vector<string> vi;//id    for(int i=0;i<N;i++){        string id;        string start,end;        cin>>id>>start>>end;        vs.push_back(transtoInt(start));        ve.push_back(transtoInt(end));        vi.push_back(id);    }    long big=0;    int big_id=0;    long small=999999999;    int small_id=0;    for(int i=0;i<N;i++){        if(small>vs[i]){            small=vs[i];            small_id=i;        }        if(big<ve[i]){            big=ve[i];            big_id=i;        }    }    cout<<vi[small_id]<<" "<<vi[big_id]<<endl;}

显示测试点都通过。