一开始很自然地想到了用sort函数,如下:
#include
#include
#include
struct stu{
char id[17];
int signIn;
int signOut;
};
int M, hh, mm, ss;
std::vector vec;
bool cmp1(const stu &a, const stu &b){
return a.signIn < b.signIn;
}
bool cmp2(const stu &a, const stu &b){
return a.signOut > b.signOut;
}
int main(){
scanf("%d", &M);
vec.resize(M);
for(int i = 0; i < M; ++i){
scanf("%s %d:%d:%d", vec[i].id, &hh, &mm, &ss);
vec[i].signIn = hh * 3600 + mm * 60 + ss;
scanf("%d:%d:%d", &hh, &mm, &ss);
vec[i].signOut = hh * 3600 + mm * 60 + ss;
}
sort(vec.begin(), vec.end(), cmp1);
printf("%s ", vec[0].id);
sort(vec.begin(), vec.end(), cmp2);
printf("%s", vec[0].id);
return 0;
}
后来一看柳婼的解法,发现用不着排序.....修改如下:
#include
#include
#include
int M, hh, mm, ss, tmp, minn, maxx;
std::string id, unlock, lock;
int main(){
std::cin >> M;
minn = 25 * 3600;
maxx = -1;
for(int i = 0; i < M; ++i){
std::cin >> id;
scanf("%d:%d:%d", &hh, &mm, &ss);
tmp = 3600 * hh + 60 * mm + ss;
if(tmp < minn){
minn = tmp;
unlock = id;
}
scanf("%d:%d:%d", &hh, &mm, &ss);
tmp = 3600 * hh + 60 * mm + ss;
if(tmp > maxx){
maxx = tmp;
lock = id;
}
}
std::cout << unlock << " " << lock;
return 0;
}
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.
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.
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.
3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40
SC3021234 CS301133