Zhejiang University has 8 campuses and a lot of gates. From each gate we can collect the in/out times and the plate numbers of the cars crossing the gate. Now with all the information available, you are supposed to tell, at any specific time point, the number of cars parking on campus, and at the end of the day find the cars that have parked for the longest time period.
Each input file contains one test case. Each case starts with two positive integers N (≤104), the number of records, and K (≤8×104) the number of queries. Then N lines follow, each gives a record in the format:
plate_number hh:mm:ss status
where plate_number
is a string of 7 English capital letters or 1-digit numbers; hh:mm:ss
represents the time point in a day by hour:minute:second, with the earliest time being 00:00:00
and the latest 23:59:59
; and status
is either in
or out
.
Note that all times will be within a single day. Each in
record is paired with the chronologically next record for the same car provided it is an out
record. Any in
records that are not paired with an out
record are ignored, as are out
records not paired with an in
record. It is guaranteed that at least one car is well paired in the input, and no car is both in
and out
at the same moment. Times are recorded using a 24-hour clock.
Then K lines of queries follow, each gives a time point in the format hh:mm:ss
. Note: the queries are given in ascending order of the times.
For each query, output in a line the total number of cars parking on campus. The last line of output is supposed to give the plate number of the car that has parked for the longest time period, and the corresponding time length. If such a car is not unique, then output all of their plate numbers in a line in alphabetical order, separated by a space.
16 7
JH007BD 18:00:01 in
ZD00001 11:30:08 out
DB8888A 13:00:00 out
ZA3Q625 23:59:50 out
ZA133CH 10:23:00 in
ZD00001 04:09:59 in
JH007BD 05:09:59 in
ZA3Q625 11:42:01 out
JH007BD 05:10:33 in
ZA3Q625 06:30:50 in
JH007BD 12:23:42 out
ZA3Q625 23:55:00 in
JH007BD 12:24:23 out
ZA133CH 17:11:22 out
JH007BD 18:07:01 out
DB8888A 06:30:50 in
05:10:00
06:30:50
11:00:00
12:23:42
14:00:00
18:00:00
23:59:00
1
4
5
2
1
0
1
JH007BD ZD00001 07:20:09
首先按照车牌号进行排序,在车牌号相同的情况下,按照进出的顺序进行排序;在这样的顺序基础上,将错误的数据删除(只有in,或者只有out,连续两个in,删除第一个in,连续两个out,删除第二个out)
错误数据删除做法:两个指针,这两个指针始终相邻,只有当满足统一车牌号in和out的时候,将两个指针的数据放入新的vector中,两个指针向后移动两个单位;否则两个指针向后移动一个单位
然后将合理的数据按照进出时间排序,因为需要查找的时间是按照时间进行排序的,所以也只需要按照时间顺序判断车辆的进出即可,然后计算停留时间最长的车辆
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