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:3 CS301111 15:30:28 17:00:10 SC3021234 08:00:00 11:25:25 CS301133 21:45:00 21:58:40Sample Output:
SC3021234 CS301133
#include<iostream>
#include<stdio.h>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct person{
string ID;
int hour,minute,second;
};
bool cmp(const person& p1,const person& p2){
if(p1.hour != p2.hour){
return p1.hour < p2.hour;
}else{
if(p1.minute != p2.minute){
return p1.minute < p2.minute;
}else{
return p1.second < p2.second;
}
}
}
int main(){
for(int n;cin>>n;){
vector<person>sign_in;
vector<person>sign_out;
for(int i = 0;i < n;i++){
person in,out;
string id ;
cin>>id;
in.ID = id;
out.ID = id;
scanf("%d:%d:%d",&in.hour,&in.minute,&in.second);
scanf("%d:%d:%d",&out.hour,&out.minute,&out.second);
sign_in.push_back(in);
sign_out.push_back(out);
}
sort(sign_in.begin(),sign_in.end(),cmp);
sort(sign_out.begin(),sign_out.end(),cmp);
cout<<sign_in[0].ID<<" "<<sign_out[sign_out.size()-1].ID<<endl;
}
return 0;
}
本文介绍了一个简单的算法问题:根据输入的一系列签到和签退记录,找出当天第一个签到(解锁门)和最后一个签退(锁门)的人。通过读取记录、按时间排序并直接选择第一个签到和最后一个签退的人员来实现。
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