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PQUEUE - Printer Queue |
The only printer in the computer science students' union is experiencing an extremely heavy workload. Sometimes there are a hundred jobs in the printer queue and you may have to wait for hours to get a single page of output.
Because some jobs are more important than others, the Hacker General has invented and implemented a simple priority system for the print job queue. Now, each job is assigned a priority between 1 and 9 (with 9 being the highest priority, and 1 being the lowest), and the printer operates as follows.
- The first job J in queue is taken from the queue.
- If there is some job in the queue with a higher priority than job J, then move J to the end of the queue without printing it.
- Otherwise, print job J (and do not put it back in the queue).
In this way, all those important muffin recipes that the Hacker General is printing get printed very quickly. Of course, those annoying term papers that others are printing may have to wait for quite some time to get printed, but that's life.
Your problem with the new policy is that it has become quite tricky to determine when your print job will actually be completed. You decide to write a program to figure this out. The program will be given the current queue (as a list of priorities) as well as the position of your job in the queue, and must then calculate how long it will take until your job is printed, assuming that no additional jobs will be added to the queue. To simplify matters, we assume that printing a job always takes exactly one minute, and that adding and removing jobs from the queue is instantaneous.
Input
One line with a positive integer: the number of test cases (at most 100). Then for each test case:
- One line with two integers n and m, where n is the number of jobs in the queue (1 ≤ n ≤ 100) and m is the position of your job (0 ≤ m ≤ n-1). The first position in the queue is number 0, the second is number 1, and so on.
- One line with n integers in the range 1 to 9, giving the priorities of the jobs in the queue. The first integer gives the priority of the first job, the second integer the priority of the second job, and so on.
Output
For each test case, print one line with a single integer; the number of minutes until your job is completely printed, assuming that no additional print jobs will arrive.
Example
Input: 3 1 0 5 4 2 1 2 3 4 6 0 1 1 9 1 1 1 Output: 1 2 5
Added by: | overwise |
Date: | 2007-10-02 |
Time limit: | 1s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS-RHINO NODEJS PERL6 VB.NET |
Resource: | ACM ICPC NWERC 2006 |
hide comments
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2015-03-04 22:03:16 cracked
My bruteforce solution got AC |
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2014-11-03 12:40:35 shiva
Very simple one using vectors |
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2014-07-07 19:29:54 fanatique
i had to comment..i spent whole day on this problem...used priority queue using heap,c++ stl queue.. simple errors: 1)same name for both global and local variables.2)compilation error when inserting "pair" in "queue". Last edit: 2014-07-07 19:31:48 |
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2014-04-03 01:11:26 Tarek Samir Sheasha
I am using a max-heap data structure to store the max-queue, and I am getting WA although all the test cases are working |
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2013-12-22 18:14:30 Rohan Phadke
interesting problem! :) easy too! |
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2013-06-23 05:12:32 Ouditchya Sinha
Nice problem on Priority Queues!! :) |
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2013-05-07 15:05:22 pawan
we are only removing High priority job when its chance will come by shifting former low pririty jobs thats why 3rd test case answer is 5. oth job is 1A. 1A 1B 9C 1D 1E 1F(higher priority present so shift to end) 1B 9C 1D 1E 1F 1A(higher priority present so shift to end) 9C 1D 1E 1F 1A 1B 1 min(remove higher priority job) 1D 1E 1F 1A 1B(now the 0th job will process after 4 min) so total time is 5 min Last edit: 2013-05-07 15:29:29 |
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2013-03-28 09:41:43 joud zouzou
can anyone explain 3rd test case? it should be 6. does the printer print the job at the position or its already printed?? EDIT: thanks pawan, i misunderstood the problem Last edit: 2013-05-15 15:44:14 |
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2013-01-27 19:40:31 ANKIT BATHLA
learn how to handle stl pair :-) |