[PAT] Class 1095 Cars on Campus (30 points)

The meaning of problems:
Enter two positive integers N and K (N <= 1e4, K <= 8e4), followed by N input data lines each comprising a string representation of three license plate number, the current time, the status of entering or leaving. Then enter K times inquiry, output current number of vehicles stuck in school. The last line of output a total residence time of the longest license plate number (ascending lexicographical output) and total residence time.

AAAAAccepted code：

// Because the day only 86,400 seconds, so you can use a car array of statistics when the vehicle enters the dwell time at +1 time, departure time -1. Then again a sweeping array sum, sum [i] = sum [i-1] + car [i], to indicate the current time how many vehicles are still staying on campus.

 1 #define HAVE_STRUCT_TIMESPEC
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 map<string,int>mp;
 5 vector<pair<string,int> >v[10007];
 6 int sum[100007],car[100007];
 7 int t[10007];
 8 map<int,string>id;
 9 vector<string>name;
10 int a[17];
11 void clear(string&s){
12     string ss;
13     swap(s,ss);
14 }
15 int main(){
16     ios::sync_with_stdio(false);
17     cin.tie(NULL);
18     cout.tie(NULL);
19     int n,k;
20     cin>>n>>k;
21     int cnt=0;
22     for(int i=1;i<=n;++i){
23         string s;
24         cin>>s;
25         string t;
26         cin>>t;
27         string f;
28         cin>>f;
29         if(!mp[s])
30             mp[s]=++cnt;
31         id[mp[s]]=s;
32         int flag=0;
33         if(f[0]=='i')
34             flag=1;
35         v[mp[s]].push_back({t,flag});
36     }
37     for(int i=1;i<=cnt;++i){
38         sort(v[i].begin(),v[i].end());
39         string now;
40         int flag=0;
41         for(int j=0;j<v[i].size();++j){
42             if(v[i][j].second==1){
43                 now=v[i][j].first;
44                 flag=1;
45             }
46             else{
47                 if(flag){
48                     string tamp=v[i][j].first;
49                     int start=(now[0]-'0')*36000+(now[1]-'0')*3600+(now[3]-'0')*600+(now[4]-'0')*60+(now[6]-'0')*10+(now[7]-' 0 ' );
50                      int over = (full view [ 0 ] - ' 0 ' ) * 36000 + (full view [ 1 ] - ' 0 ' ) * 3600 + (full view [ 3 ] - ' 0 ' ) * 600 + (full view [ 4 ] - ' 0 ' ) * 60 + (full view [ 6 ] - ' 0 ' ) * 10 + (full view [ 7 ] - ' 0 ' );
51                     ++car[start];
52                     --car[over];
53                     flag=0;
54                     clear(now);
55                     t[i]+=over-start;
56                 }
57             }
58         }
59     }
60     sum[0]+=car[0];
61     for(int i=0;i<86400;++i)
62         sum[i]=sum[i-1]+car[i];
63     for(int i=1;i<=k;++i){
64         string tamp;
65         cin>>tamp;
66         int tt=(tamp[0]-'0')*36000+(tamp[1]-'0')*3600+(tamp[3]-'0')*600+(tamp[4]-'0')*60+(tamp[6]-'0')*10+(tamp[7]-'0');
67         cout<<sum[tt]<<"\n";
68     }
69     int mx=0;
70     for(int i=1;i<=cnt;++i)
71         mx=max(mx,t[i]);
72     for(int i=1;i<=cnt;++i)
73         if(mx==t[i])
74             name.push_back(id[i]);
75     sort(name.begin(),name.end());
76     for(int i=0;i<name.size();++i)
77         cout<<name[i]<<" ";
78     a[1]=mx/36000;
79     mx%=36000;
80     a[2]=mx/3600;
81     mx%=3600;
82     a[3]=mx/600;
83     mx%=600;
84     a[4]=mx/60;
85     mx%=60;
86     a[5]=mx/10;
87     mx%=10;
88     a[6]=mx;
89     cout<<a[1]<<a[2]<<":"<<a[3]<<a[4]<<":"<<a[5]<<a[6];
90     return 0;
91 }