1095 Cars on Campus (30分)--PAT甲级真题，测试点1,4

hejiang 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
(≤8×10​⁴​​ ) 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

题目大意：给出N条车辆进出入的信息；之后有K个查询，查询某个时刻学校里面车量的数量；最后输出在校园里停车时间最长的车牌，如果有多个，则按照字母递增顺序输出；

分析：
这个题目的难点，还有坑点；是输入当中会有无用信息；比如 JH007BD 05:09:59 in； JH007BD 05:10:33 in；这两条信息，就只有后面一条是有用的输入，因为没有与 JH007BD 05:09:59 in相配对的out;但是JH007BD 12:23:42 outt与靠他最近的in JH007BD 05:10:33 in刚好匹配；同理，JH007BD 12:24:23 out也是一条无用信息；可以把输入的信息先按照车牌排序，然后按照时间排序，找出前后匹配的记录，删除无用记录；

测试点1，4错误，但是其他结果都正确的话，建议检查一下计算学校里车数量的那一部分，我之前错了这两个测试点，检查之后发现是这部分写错了；

#include
#include
#include 
#include
#include  
using namespace std; 
int cars[3600 * 24 + 1] = { 0 };
struct Info {
	string ID, status;
	int t;
	bool operator<(Info& A) { return ID == A.ID ? t < A.t : ID < A.ID; }
};
bool cmp(Info& A, Info& B) { return A.t < B.t; }
vector<Info>Vin;
map<string, vector<Info>>CarInfo;//每辆车的信息集合
map<string,int>Time;//每辆车停车的时间
int Total = 0;//车库的车辆数目
int main(){
	int N, K, lastT = 0;
	scanf("%d%d", &N, &K); 
	Vin.resize(N);
	for (int i = 0; i < N; i++) {//输入 
		int h, m, s;
		string status;
		cin >> Vin[i].ID;
		scanf("%d:%d:%d", &h, &m, &s);
		cin >> Vin[i].status;
		Vin[i].t = 3600 * h + 60 * m + s; 
	} 
	sort(Vin.begin(), Vin.end());
	vector<Info>Vtmp;
	for (int i =1; i<N; i++) {//筛选有用记录  
		if (Vin[i - 1].ID== Vin[i].ID&&Vin[i-1].status == "in"&&Vin[i].status == "out") { 
			Time[Vin[i - 1].ID] += Vin[i].t - Vin[i - 1].t;
			Vtmp.push_back(Vin[i - 1]);
			Vtmp.push_back(Vin[i]);
		}  
	}
	Vin.clear();
	Vin = Vtmp;
	sort(Vin.begin(), Vin.end(), cmp);
	for (int i = 0; i < Vin.size(); i++){ 
		int t = Vin[i].t;
		if (Vin[i].status == "in") { 
			if (t == lastT) {//有多个车辆同时进的情况
				Total++;
				continue; 
			} 
			fill(cars + lastT, cars + t, Total++);
			lastT = t;
		}
		else {
			if (t == lastT) {//有多个车辆同时出的情况
				Total--;
				continue;
			}
			fill(cars + lastT, cars + t, Total--);
			lastT = t;
		}  
	}  
	fill(cars + lastT, cars + 3600 * 24 + 1, Total); 
	for (int i = 0; i < K; i++){
		int h, m, s, t;
		scanf("%d:%d:%d", &h, &m, &s);
		t = 3600 * h + 60 * m + s;
		printf("%d\n", cars[t]);
	}  
	int maxT = -1;
	vector<string>V;
	for (auto it = Time.begin(); it != Time.end(); it++) {   
		if (it->second > maxT) {
			maxT = it->second;
			V.clear();
			V.push_back(it->first);
		}
		else if (it->second == maxT)
			V.push_back(it->first); 
	}
	sort(V.begin(), V.end());
	for (int i =0; i<V.size(); i++)
			printf("%s ", V[i].c_str()); 
	printf("%02d:%02d:%02d", maxT / 3600, (maxT % 3600) / 60, maxT % 60);
	return 0;
}