1095 cars on campus (30 points)

The main idea of the topic ： Give the entry and exit time of each license plate , every last in Only with the next one with the same number out matching , If there's no match in or out Is ignored , Then give the corresponding inquiry , Find out how many cars stay in a certain period of time , And find the longest time to stay .

Ideas ：

1. First, read all the data and sort by license plate number and time .( Convert time into seconds )

2. Comparing the two , Deal with legal license plates , And use map Save the total stay time of each license plate , Record the maximum .

3. Deal with inquiries , The number of vehicles staying in each time period can be pretreated ( The prefix and , After testing, there is no prefix and no timeout )

4. The last step is to traverse map Output the license plate with the same dwell time

If the vehicle that stays for the longest time is not unique , Output in dictionary order , It can be used directly map save , Guaranteed dictionary order .

#include <iostream>
#include <algorithm>
#include <vector>
#include <map>
using namespace std;
// Convert time into seconds 
struct node
{
	string s;
	int time,flag;
	bool operator < (const node&W)
	{	
        if(s != W.s) return s < W.s;// Put the same brand together 
		return time < W.time;
	}
}e[10010];
int pre[10010];
map<string,int>mp;
int main()
{
	int n,k,maxtime = 0;
	cin >> n >> k;
	for(int i = 0; i < n; i ++)
	{    
	    char zt[10];int a,b,c;
        cin >> e[i].s;
		scanf("%d:%d:%d %s",&a,&b,&c,zt);
		e[i].time = a * 3600 + b * 60 + c;
		if(zt[0] == 'i') e[i].flag = 1;
		else e[i].flag = -1;
	}
	sort(e,e + n);
	vector<node>ans;
	for(int i = 0 ; i < n - 1; i ++)
	{
		if(e[i].s == e[i + 1].s && e[i].flag == 1 && e[i + 1].flag == -1)
		{
			ans.push_back(e[i]),ans.push_back(e[i + 1]);
			mp[e[i].s] +=(e[i + 1].time - e[i].time);
			maxtime = max(maxtime,mp[e[i].s]);
		}
	}
	sort(ans.begin(),ans.end(),[](node a,node b)
         {
             return a.time < b.time;
         });
	for(int i = 0 ; i < ans.size(); i ++)
	{
		if(i == 0) pre[i] = ans[i].flag;
		else pre[i] = pre[i - 1] + ans[i].flag;// The prefix and 
	}
	int last = 0;
	while(k --)
	{
	   int a,b,c;
	   scanf("%d:%d:%d",&a,&b,&c);
	   int t = a * 3600 + b * 60 + c,cnt = 0;
	   for(int i = last; i < ans.size(); i ++)
	   {
		   if(ans[i].time > t)
		   {
			   printf("%d\n",pre[i - 1]);
			   break;
		   }
		   else if (i == ans.size() - 1) printf("%d\n",pre[i]);
		   last = i;
	   }
	}
	for(auto it : mp)
	{
		if(it.second == maxtime) printf("%s ",it.first.c_str());
	}
	printf("%02d:%02d:%02d",maxtime / 3600, maxtime % 3600 / 60, maxtime % 60);
	return 0;
}