1095 Cars on Campus

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配对要求是，如果一个车多次进入未出，取最后一个值；如果一个车多次out未进入，取第一个值。
注意：一个车可能出入校园好多次，停车的时间应该取之和

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <string.h>
#include <vector>
#include <unordered_map>
using namespace std;
const int N=10010;
const int MAXT=24*3600;
int n,k,hh,mm,ss,flag,num,maxv;
string p,st;
unordered_map<string,int>plates;
unordered_map<int,string>revplates;
vector<string>ansstr;
struct Car{
	int tot,flag;
	bool operator<(Car x)const{
		return tot<x.tot;
	}
};
vector<Car>cars[N];
int timeList[MAXT];
int main() {
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;++i){
		cin>>p;
		if(!plates.count(p)) plates[p]=++num;
		revplates[plates[p]]=p;
		scanf("%d:%d:%d",&hh,&mm,&ss);
		int tot=hh*3600+mm*60+ss;
		cin>>st;
		if(st=="in") flag=1;
		else flag=0;
		cars[plates[p]].push_back({tot,flag});
	}
	for(int i=1;i<=num;++i){
		sort(cars[i].begin(),cars[i].end());
		int j=0,span=0;
		while(j<cars[i].size()){
			while(!cars[i][j].flag&&j<cars[i].size()) j++;
			if(j>=cars[i].size()) break;
			while(cars[i][j].flag&&j<cars[i].size()) j++;
			if(j>=cars[i].size()) break;
			//cars[i][j-1].flag==1,cars[i][j].flag==0
			timeList[cars[i][j-1].tot]++;
			timeList[cars[i][j].tot]--;
			span+=cars[i][j].tot-cars[i][j-1].tot;
			j++;
		}
		if(span>maxv){
			ansstr.clear();
			ansstr.push_back(revplates[i]);
			maxv=span;
		}else if(span==maxv){
			ansstr.push_back(revplates[i]);
		}
	}
	for(int i=1;i<MAXT;++i)
		timeList[i]+=timeList[i-1];
	sort(ansstr.begin(),ansstr.end());
	while(k--){
		scanf("%d:%d:%d",&hh,&mm,&ss);
		int tot=hh*3600+mm*60+ss;
		printf("%d\n",timeList[tot]);
	}
	for(int i=0;i<ansstr.size();++i)
		cout<<ansstr[i]<<" ";
	printf("%02d:%02d:%02d\n",maxv/3600,(maxv/60)%60,maxv%60);
	return 0;
}