1017. Queueing at Bank (25)

Suppose a bank has K windows open for service. There was a yellow line in front of the windows which devides the waiting area into the parts. All the customers has to wait for line behind the Yellow Line, until it's his/her turn to being served and there is a window Available. It is assumed this no window can be occupied by a single customer for more than 1 hour.

Now given the arriving time T and the processing time P of each customer, you is supposed to tell the average waiting Tim E of all the customers.

Input Specification:

Each input file contains the one test case. The first line contains 2 numbers:n (<=10000)-the total number of customers, and K (<=100)-The N Umber of Windows. Then N lines follow, each contains 2 times:hh:mm:ss-the arriving time, and p-the processing time in minutes of a Cust Omer. Here HH was in the range [XX,], MM and SS are both in [00, 59]. It is assumed, that no, customers arrives at the same time.

Notice the bank opens from 17:00. Anyone arrives early'll has to wait for line till, and anyone comes too late (at or after 17:00:01) won't be SE RVed nor counted into the average.

Output Specification:

For each test case, print on one line the average waiting time of the "all" the customers, in minutes and accurate up to 1 DECIM Al Place.

Sample Input:

7 307:55:00 1617:00:01 207:59:59 1508:01:00 6008:00:00 3008:00:02 208:03:00 10

Sample Output:

8.2

1#include <stdio.h>2#include <vector>3#include <algorithm>4 using namespacestd;5 structcus6 {7     intArrive_time,cost;8 };9 structwinTen { OneWin (): Win_time (8* -* -){} A     intWin_time; - }; -  theWin wins[ $]; -  - BOOLcmparrive (cus a,cus b) - { +     returnA.arrive_time <B.arrive_time; - } + intMain () A { at     intN,w_num,hh,mm,ss; - cus Ctem; -Vector<cus>VV; -scanf"%d%d",&n,&w_num); -      for(inti =0; I < n;++i) -     { inscanf"%d:%d:%d%d",&hh,&mm,&ss,&ctem.cost); -         if(Ctem.cost > -) toCtem.cost = -; +Ctem.cost *= -; -Ctem.arrive_time = hh* -* -+ mm* -+SS; the         if(Ctem.arrive_time <= -* -* -) * Vv.push_back (ctem); $     }Panax Notoginseng sort (Vv.begin (), Vv.end (), cmparrive); -     intCNT =0; the     Doublesum =0.0; +n =vv.size (); A      while(CNT <N) the     { +         intMIN =99999999; -         intMinindex =-1; $          for(inti =0; I < w_num;++i) $         { -             if(MIN >wins[i].win_time) -             { theMIN =Wins[i].win_time; -Minindex =i;Wuyi             } the         } -         if(MIN >vv[cnt].arrive_time) Wu         { -Sum + = (MIN-vv[cnt].arrive_time); AboutWins[minindex].win_time + =Vv[cnt].cost; $++CNT; -         } -         Else -         { A              for(inti =0; I < w_num;++i) +             { the                 if(Vv[cnt].arrive_time >=wins[i].win_time) -                 { $Minindex =i; the                      Break; the                 } the             } theWins[minindex].win_time = Vv[cnt].arrive_time +Vv[cnt].cost; -++CNT; in         } the     } the     if(n = =0) Aboutprintf"0.0\n"); the     Else theprintf"%.1lf\n", sum/60.0/n); the     return 0; +}

1017. Queueing at Bank (25)