PAT_A 1095 Cars on Campus

PAT_A 1095 Cars on Campus

分析

题目要求对停车时间进行分析，要求查询任一时刻（秒）的停车数量，且需要输出停车时间最长车的车牌号，若有多辆车并列，则按字母升序输出。本题对性能的要求较高，在提交时内存、耗时都是压线过的...可以说侥幸勉强的通过了

• 在读入信息时，使用map查询车的索引。

• 若所有车按总时间排序会超时，因此只找出停车最长的按照字母顺序排序

• 查询时间使用了表示全天任意秒的时刻的数组

• 题目中输入的时间是一天内的

• 提交的代码中有部分范围数据是试出来的，为了保证内存符合要求...

• 查询时间是递增的，且时间应一开始就使用秒来存，改进后应该能够提高性能

  于是又找到了一份质量较高的代码，以供学习参考

题目的描述

Zhejiang 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 \(N\ (≤10^4)\), 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

较低质量的AC的代码

#include <bits/stdc++.h>

using namespace std;

const static int MAXN = 10001,MAXK = 80001,DAYTIME=24*60*60+1;

array<int,DAYTIME> sslist;
void ssadd(int s,int e){
    for(int i=s;i<e;i++){
        sslist[i]+=1;
    }
}

class times{
    public:
    int hh,mm,ss;
    bool of;//in true; out false
    times(){
        
    }
    int operator-(times t){
        return 60*60*hh+60*mm+ss-60*60*t.hh-60*t.mm-t.ss;
    }
    int tos(){
        return 60*60*hh+60*mm+ss;
    }
};
bool cmp1(times a,times b){
    if(a.hh==b.hh){
        return a.mm ==b.mm ?
            a.ss < b.ss
            : a.mm<b.mm;
    }
    return a.hh < b.hh;
}
class Car{
    public:
    char plate_number[10];
    array<times,300> timeslist;
    int tot,tl;
    Car(){
        tl = 0;
        tot = 0;
    }
    void calculate(){
        sort(timeslist.begin(),timeslist.begin()+tl,cmp1);
        for(int i=1;i<tl;i++){
            if(timeslist[i-1].of&&!timeslist[i].of){
                tot += timeslist[i]-timeslist[i-1];
                ssadd(timeslist[i-1].tos(),timeslist[i].tos());
                i+=1;
            }
        }
    }
    char* tottostring(){
        char *res=(char*)malloc(sizeof(char)*10);
        int h= tot/60/60,m=tot%(60*60)/60,s = tot%(60*60)%60;
        sprintf(res,"%02d:%02d:%02d",h,m,s);
        return res;
    }
};
bool cmp2(Car a,Car b){
//     return a.tot == b.tot ? strcmp(a.plate_number,b.plate_number) < 0 : a.tot>b.tot;
    return strcmp(a.plate_number,b.plate_number)<0;
}
array<Car,MAXN> cars;
array<Car,5000> longest;
int main(){
    int N,K,cl=0,ll=0;
    sslist.fill(0);
    scanf("%d%d",&N,&K);
    map<string,int> ci;
    for(int i=0;i<N;i++){
        char t[10];
//         bool f=false;
        scanf("%s",t);
        string t2 = t;
//         for(int j=0;j<cl;j++)
        if(ci.count(t2)!=0){int j=ci[t2];
//             if(0==strcmp(t,cars[j].plate_number))
            {
                strcpy(cars[j].plate_number,t);
                scanf("%d:%d:%d",
                      &cars[j].timeslist[cars[j].tl].hh,
                      &cars[j].timeslist[cars[j].tl].mm,
                      &cars[j].timeslist[cars[j].tl].ss
                     );
                scanf("%s",t);
                if(0==strcmp(t,"in")){
                    cars[j].timeslist[cars[j].tl].of=true;
                }
                else{
                    cars[j].timeslist[cars[j].tl].of=false;
                }
                cars[j].tl+=1;
//                 f = true;
//                 break;
            }
        }
//         if(!f)
        else{
            ci[t2]=cl;
            strcpy(cars[cl].plate_number,t);
            scanf("%d:%d:%d",
                  &cars[cl].timeslist[cars[cl].tl].hh,
                  &cars[cl].timeslist[cars[cl].tl].mm,
                  &cars[cl].timeslist[cars[cl].tl].ss
                 );
            scanf("%s",t);
            if(0==strcmp(t,"in")){
                cars[cl].timeslist[cars[cl].tl].of=true;
            }
            else{
                cars[cl].timeslist[cars[cl].tl].of=false;
            }
            cars[cl].tl+=1;
            cl+=1;
        }
    }
    int m = cars[0].tot;
    for(int i=0;i<cl;i++){
        cars[i].calculate();
        if(cars[i].tot>m){
            m = cars[i].tot;
        }
//         cout<<cars[i].plate_number<<endl;
//         for(int j=0;j<cars[i].tl;j++){
//             cout<<cars[i].timeslist[j].hh<<":"<<cars[i].timeslist[j].mm<<':'<<cars[i].timeslist[j].ss<<' '<<cars[i].timeslist[j].of<<endl;
//         }
//         cout<<cars[i].tot<<' '<<endl;
//         printf("%s\n",cars[i].tottostring());
    }
//     sort(cars.begin(),cars.begin()+cl,cmp2);
    for(int i=0;i<cl;i++){
        if(m==cars[i].tot){
            longest[ll++]=cars[i];
        }
    }
    sort(longest.begin(),longest.begin()+ll,cmp2);
    for(int i=0;i<K;i++){
        int h,m,s;
        scanf("%d:%d:%d",&h,&m,&s);
        printf("%d\n",sslist[60*60*h+60*m+s]);
    }

    
    for(int i=0;i<ll;i++){
        printf("%s ",longest[i].plate_number);
    }
    printf("%s\n",longest[0].tottostring());
    return 0;
}

较高质量的AC的代码

#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<algorithm>
using namespace std;
const int maxn = 10010;
struct Car{
    char id[8];
    int time;
    char status[4];
}all[maxn],valid[maxn];
int num=0;
map<string,int> parkTime;
int timeToInt(int hh,int mm,int ss){
    return hh*3600+mm*60+ss;
}
bool cmpByIdAndTime(Car a,Car b){
    if(strcmp(a.id,b.id)) return strcmp(a.id,b.id)<0;
    else return a.time<b.time;
}
bool cmpByTime(Car a,Car b){
    return a.time<b.time;
}
int main(){
    int n,k,hh,mm,ss;
    scanf("%d%d",&n,&k);
    for(int i=0;i<n;i++){
        scanf("%s %d:%d:%d %s",all[i].id,&hh,&mm,&ss,all[i].status);
        all[i].time = timeToInt(hh,mm,ss);
    }
    sort(all,all+n,cmpByIdAndTime);
    int maxTime = -1;
    for(int i=0;i<n-1;i++){
        if(!strcmp(all[i].id,all[i+1].id)&&
          !strcmp(all[i].status,"in")&&
           !strcmp(all[i+1].status,"out")){
            valid[num++] = all[i];
            valid[num++] = all[i+1];
            int inTime = all[i+1].time - all[i].time;
            if(parkTime.count(all[i].id)==0){
                parkTime[all[i].id] = 0;
            }
            parkTime[all[i].id] += inTime;
            maxTime = max(maxTime,parkTime[all[i].id]);
        }
    }
    sort(valid,valid+num,cmpByTime);
    int now = 0,numCar = 0;
    for(int i=0;i<k;i++){
        scanf("%d:%d:%d",&hh,&mm,&ss);
        int time = timeToInt(hh,mm,ss);
        while(now<num&&valid[now].time<=time){
            if(!strcmp(valid[now].status,"in"))numCar++;
            else numCar--;
            now++;
        }
        printf("%d\n",numCar);
    }
    map<string,int>::iterator it;
    for(it = parkTime.begin();it!=parkTime.end();it++){
        if(it->second == maxTime){
            printf("%s ",it->first.c_str());
        }
    }
    printf("%02d:%02d:%02d\n",maxTime/3600,maxTime%3600/60,maxTime%60);
    return 0;
}