PAT (Advanced Level) Practise - 1095. Cars on Campus (30)

 

http://www.patest.cn/contests/pat-a-practise/1095 

Zhejiang University has 6 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 (<= 10000), the number of records, and K (<= 80000) 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

这道题是2015考研机试前的那个PAT的D题 http://www.patest.cn/contests/pat-a-101-125-1-2015-03-14

那次PAT据说比机试简单。。。于是很多高分大神申请了免机试。。。性价比超高。。。。

我表示很忧伤。。。虽然我对自己机试成绩还算满意，但是还是有点感慨。。。如果参加这次PAT 说不定分数更好。。。嗯，做梦的感觉好好好哦。。。

T_T  并不能轻易的承认 这貌似是我在pat上扯过的最长的代码了。。。不过好消息是A级我才做三分之一不到 哈哈哈哈 说不定有更长的

#include<cstdio>
#include<cstring>
struct carrecord
{
       long long carid;
       int  second;
       int  status;  // 1-in  0-out
}records[10001];
int recordsnum=0,queriesnum=0,time[10000][2]={0};


long long  str2int(char *str)//一个大于 （26字母+10数字）的数值即可 用于保持比较时有效的字典序列 
{
    long long  num=0,istr=0;
    while(str[istr]) num=num*40+(('0'<=str[istr]&&str[istr]<='9')?str[istr]-'0':str[istr]-'A'+10),istr++;  
    return num;
}

void outlook(long long  num)//一条记录输出 
{
     char car[8]="0000000";
     int istr=6,temp=0;
     car[7]='\0';
     while(istr>=0)//int2str  恢复字符串名字 
     {
          temp=num%40;
          if(temp>=10) car[istr]=temp+'A'-10;
          else car[istr]=temp+'0'; 
          num/=40,istr--;
     }
     printf("%s ",car);
}


//基于汽车牌号（以转为对应序列的数值）和进出时间的快排 
void QS(int low,int high,const int iqs) //1-carid  0-time
{
     int l=low,h=high,second=records[l].second,status=records[l].status;
     long long  carid=records[l].carid;
     
     while(l<h)
     {
         if(iqs) while( l<h && (carid<records[h].carid ||(carid==records[h].carid && second<=records[h].second))) h--;
         else while( l<h && (second<records[h].second ||(  carid<=records[h].carid && second==records[h].second  ))) h--;         
         if(l<h)
         {
                records[l].carid=records[h].carid;
                records[l].second=records[h].second;
                records[l].status=records[h].status;
                
                records[h].carid=carid;
                records[h].second=second;
                records[h].status=status;
         }
               
         if(iqs) while( l<h && (carid>records[l].carid||(carid==records[l].carid&& second>=records[l].second))) l++;
         else while( l<h && (second>records[l].second ||(carid>=records[l].carid&& second==records[l].second))) l++;
         if(l<h)
         {
                records[h].carid=records[l].carid;
                records[h].second=records[l].second;
                records[h].status=records[l].status;
                
                records[l].carid=carid;
                records[l].second=second;
                records[l].status=status;
         }
     }     
     if(low+1<l) QS(low,l-1,iqs);
     if(h+1<high) QS(h+1,high,iqs);
}


int clean(int len,int in,int out)
{
    int l=len,h=in;
    records[l].carid=records[h].carid;
    records[l].second=records[h].second;
    records[l].status=records[h].status; 
    
    l=l+1,h=out;
    records[l].carid=records[h].carid;
    records[l].second=records[h].second;
    records[l].status=records[h].status; 
                      
    return len+2;
}


int main()
{
    scanf("%d%d",&recordsnum,&queriesnum);
    
    char carname[8],status[4];
    int hh,mm,ss;
    for(int i=1;i<=recordsnum;i++)//从1开始是为了方便下面记录配对时直接移动即可 而不需temp 
    {
            scanf("%s %d:%d:%d %s",carname,&hh,&mm,&ss,status);
            records[i].carid=str2int(carname);//  比较、移动、复制等，操作简单，节省时间 
            records[i].second=ss+60*(mm+60*hh);
            records[i].status=(0==strcmp(status,"in")?1:0);
    }
    QS(1,recordsnum,1);// 基于汽车牌号（以转为对应序列的数值）的QS 
    
    
    int num=1,len=0,flagin=-1,flagout=-1;
    long long carid=0;
    while(num<=recordsnum)// Each "in" record is paired with the chronologically next record for the same car provided it is an "out" record
    {
            flagin=-1,flagout=-1,carid=records[num].carid;
            while(carid==records[num].carid && num<=recordsnum) // Any "in" records that are not paired with an "out" record are ignored, as are "out" records not paired with an "in" record.
            {
                    if(1==records[num].status) flagin=num,flagout=-1;
                    else if(-1==flagin) flagin=-1,flagout=-1;
                    else len=clean(len,flagin,num),flagin=-1,flagout=-1;                    
                    num++;
            }
    }
    recordsnum=len;
    QS(0,recordsnum-1,0);
        
    num=0;
    int istr=0,carnum=0,notfirst=1;
    if(recordsnum) notfirst=0;
    while(num<queriesnum)   //For each query, output in a line the total number of cars parking on campus
    {
            scanf("%d:%d:%d",&hh,&mm,&ss);
            len=ss+60*(mm+60*hh);
            while(len>=records[istr].second && istr<recordsnum)
            {
                if(records[istr].status) carnum++;
                else carnum--;
                istr++;
            }
            if(num<queriesnum-notfirst) printf("%d\n",carnum);
            else printf("%d",carnum);
            num++;
    }
    
    
    QS(0,recordsnum-1,1);
    num=0,istr=0,flagout=0;
    while(num<recordsnum)   //the longest time period  parked for
    {
            istr=0,carid=records[num].carid;
            while(carid==records[num].carid && num<recordsnum)
            {
                    istr+=records[num+1].second-records[num].second;
                    records[num+1].second=-1,records[num].second=-1;                                                       
                    num+=2;
            }
            records[num-1].second=istr;
            if(istr>flagout) flagout=istr;
    }
    
    for(int i=0;i<recordsnum;i++) 
            if(records[i].second==flagout) outlook(records[i].carid);  //give the plate number of the car that has parked for the longest time period
    printf("%02d:%02d:%02d",flagout/3600,(flagout/60)%60,flagout%60);  //and the corresponding time length
    return 0;
    
}