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 21:58:40
Sample Output:
SC3021234 CS301133
题目分析：给定数据排序 ，直接利用vector对结构体排序 将比较函数重写就可以了
注意sort中比较函数 返回true是不交换的 鉴于现在我水平较低 之后会尝试分析 源码的(大概
1 #include<iostream> 2 #include<string> 3 #include<stdlib.h> 4 #include<vector> 5 #include<algorithm> 6 using namespace std; 7 #define MaxNum 1 8 typedef struct Node { 9 string ID_numebr; 10 string Sign_in_time; 11 string Sign_out_time; 12 }S; 13 typedef struct Time { 14 int H; 15 int M; 16 int S; 17 }T; 18 T t1, t2; 19 int Atoi(string str, int i, int j) 20 { 21 int num = 0; 22 while (i<j) 23 { 24 num = num * 10 + str[i] - '0'; 25 i++; 26 } 27 return num; 28 } 29 bool comp2(const S& a, const S& b) 30 { 31 32 t1.H = Atoi(a.Sign_out_time, 0, 2); 33 t1.M = Atoi(a.Sign_out_time, 3, 5); 34 t1.S = Atoi(a.Sign_out_time, 6, 8); 35 t2.H = Atoi(b.Sign_out_time, 0, 2); 36 t2.M = Atoi(b.Sign_out_time, 3, 5); 37 t2.S = Atoi(b.Sign_out_time, 6, 8); 38 if (t1.H > t2.H) 39 return true; 40 else if (t1.H == t2.H && t1.M > t2.M) 41 return true; 42 else if (t1.M == t2.M && t1.S > t2.S) 43 return true; 44 else 45 return false; 46 } 47 48 bool comp1(const S& a, const S& b) 49 { 50 51 t1.H = Atoi(a.Sign_in_time, 0, 2); 52 t1.M = Atoi(a.Sign_in_time, 3, 5); 53 t1.S = Atoi(a.Sign_in_time, 6, 8); 54 t2.H = Atoi(b.Sign_in_time, 0, 2); 55 t2.M = Atoi(b.Sign_in_time, 3, 5); 56 t2.S = Atoi(b.Sign_in_time, 6, 8); 57 if (t1.H < t2.H) 58 return true; 59 else if (t1.H == t2.H && t1.M < t2.M) 60 return true; 61 else if (t1.M == t2.M && t1.S < t2.S) 62 return true; 63 else 64 return false; 65 } 66 67 int main() 68 { 69 vector<Node> N; 70 S s; 71 int n; 72 cin >> n; 73 for (int i = 0; i < n; i++) 74 { 75 cin >> s.ID_numebr >> s.Sign_in_time >> s.Sign_out_time; 76 N.push_back(s); 77 } 78 sort(N.begin(), N.end(), comp1); 79 cout << N[0].ID_numebr<<" "; 80 sort(N.begin(), N.end(), comp2); 81 cout << N[0].ID_numebr; 82 return 0; 83 }