分析：要求输出某个时刻停车场的人数，若in和out不匹配则本条记录无效，如in之后还是in则该辆车是无效的。本题防止超时的关键就是给的query是递增的，就不用每次都从0开始遍历。
#include <iostream>
#include<cstring>
#include<vector>
#include<stdio.h>
#include<queue>
#include<math.h>
#include<stack>
#include<algorithm>
#include<map>
#include<set>
#include<iostream>
#include<stdlib.h>
#include<string.h>
using namespace std;
#define MAX 999999999
typedef long long ll;
int n,m,k;
struct Node{
string plate;
int time;
string status;
int flag;
int yes;
}node[100001];
map<string,int>ma;
int cmp(Node x, Node y)
{
if(x.plate == y.plate)
return x.time < y.time;
return x.plate < y.plate;
}
int cmp1(Node x, Node y)
{
return x.time < y.time;
}
int main()
{
cin>>n>>k;
for(int i = 0; i <n; i++)
{
string x,y;
int a,b,c;
cin>>x;
scanf("%d:%d:%d",&a,&b,&c);
cin>>y;
node[i].plate = x;
node[i].status = y;
node[i].time = a * 3600 + b * 60 + c;
node[i].flag = y[0]=='i'?0:1;
}
sort(node,node+n,cmp);
int mmax = -1;
for(int i = 0 ; i < n - 1 ;i ++)
{
if(node[i].plate == node[i+1].plate && node[i].flag == 0 && node[i+1].flag == 1){
ma[node[i].plate] += node[i+1].time - node[i].time;
mmax = max(mmax, ma[node[i].plate]);
node[i].yes = node[i+1].yes = 1;
}
}
sort(node,node+n,cmp1);
int j = 0,cnt = 0;
for(int i = 0 ; i < k ; i++)
{
int a,b,c;
scanf("%d:%d:%d",&a,&b,&c);
int t = a * 3600 + b * 60 + c;
for(; j < n && node[j].time <= t;j ++)
{
if(node[j].yes == 1 && node[j].status == "in"){
cnt++;
}
if(node[j].yes == 1 && node[j].status == "out"){
cnt--;
}
}
cout<<cnt<<endl;
}
for(auto it = ma.begin();it!=ma.end();it++)
{
if(it->second == mmax)
cout<<it->first<<" ";
}
printf("%02d:%02d:%02d",mmax/3600,mmax%3600/60,mmax%60);
return 0;
}