1006 Sign In and Sign Out(25) - PAT甲级

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

Solution:

时间可以存成字符数组, 用strcmp进行比较, strcmp(time1, time2) < 0, time1比time2早。
只要进行一次升序和一次降序, 就可以得出最早和最晚的时间, 对应的id, 因此存在一个struct中.

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;

struct Node {
  char id[20];
  char in_time[15];
  char out_time[15];
};

bool cmp1(Node n1, Node n2) {
  return strcmp(n1.in_time, n2.in_time) < 0;
}

bool cmp2(Node n1, Node n2) {
  return strcmp(n1.out_time, n2.out_time) > 0;
}

vector<Node> v;

int main() {
  int M; scanf("%d", &M);
  for (int i = 0; i < M; i++) {
    Node n;
    scanf("%s %s %s", n.id, n.in_time, n.out_time);
    v.push_back(n);
  }
  sort(v.begin(), v.end(), cmp1);
  printf("%s ", v[0].id);
  sort(v.begin(), v.end(), cmp2);
  printf("%s", v[0].id);
  return 0;
}