PAT甲级真题 1095 Cars on Campus (30分) C++实现（结构体排序，匹配停车场in、out信息）

题目

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.
Input Specification:
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.
Output Specification:
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.

Sample Input:

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

Sample Output:

1
4
5
2
1
0
1
JH007BD ZD00001 07:20:09

思路

用结构体存储停车信息，将时间统一转化为秒数存储。主要步骤如下：

1. 筛选有效信息

读入所有信息后按时间排序，然后筛选有效信息。

顺序遍历所有信息，对于每个信息是“in”的，从该位置向后搜索同一个车牌的信息：

1. 若找到一个“out”则成功匹配，将两条信息都加入新的数组中，将匹配的"out"标记为空“”，表示已匹配过了防止重复匹配；
2. 若找到一个同名的信息不是“out”（又一个“in”），则第一个"in"无效，直接跳过。

2. 统计最大停车时长

上一步匹配in、out的过程中，顺便统计每个车牌的总停车时间（同一个车牌不同时段要累计的，这点容易忽略），用map<string, int>存储；顺便再统计最长的时间，及相应车牌（用vector< string > 存储）。

3. 查询时间节点内停车数目

将有效信息按时间排序。顺序遍历，在给定时间段内，每遇到一个in就num+1，遇到一个out就num-1。

由于查询信息是时间升序的，可利用上一次的数量（num）和遍历位置（j），减少复杂度。

代码

#include <iostream> 
#include <vector> 
#include <unordered_map> 
#include <algorithm> 
using namespace std; 

struct Info{
   
   
    string id;
    int time;
    string state;
};

bool cmp