GCJ1C09C - Bribe the Prisoners

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In a kingdom there are prison cells (numbered 1 to P) built to form a straight line segment. Cells number i and i+1 are adjacent, and prisoners in adjacent cells are called "neighbours." A wall with a window separates adjacent cells, and neighbours can communicate through that window.

All prisoners live in peace until a prisoner is released. When that happens, the released prisoner's neighbours find out, and each communicates this to his other neighbour. That prisoner passes it on to his other neighbour, and so on until they reach a prisoner with no other neighbour (because he is in cell 1, or in cell P, or the other adjacent cell is empty). A prisoner who discovers that another prisoner has been released will angrily break everything in his cell, unless he is bribed with a gold coin. So, after releasing a prisoner in cell A, all prisoners housed on either side of cell A - until cell 1, cell P or an empty cell - need to be bribed.

Assume that each prison cell is initially occupied by exactly one prisoner, and that only one prisoner can be released per day. Given the list of Q prisoners to be released in Q days, find the minimum total number of gold coins needed as bribes if the prisoners may be released in any order.

Note that each bribe only has an effect for one day. If a prisoner who was bribed yesterday hears about another released prisoner today, then he needs to be bribed again.


The first line of input gives the number of cases, N. N test cases follow. Each case consists of 2 lines. The first line is formatted as


where P is the number of prison cells and Q is the number of prisoners to be released.
This will be followed by a line with Q distinct cell numbers (of the prisoners to be released), space separated, sorted in ascending order.


For each test case, output one line in the format

Case #X: C
where X is the case number, starting from 1, and C is the minimum number of gold coins needed as bribes.


1 ≤ N ≤ 100
Each cell number is between 1 and P, inclusive.

Large dataset

1 ≤ P ≤ 10000
1 ≤ Q ≤ 100



8 1
20 3
3 6 14
Case #1: 7
Case #2: 35


In the second sample case, you first release the person in cell 14, then cell 6, then cell 3. The number of gold coins needed is 19 + 12 + 4 = 35. If you instead release the person in cell 6 first, the cost will be 19 + 4 + 13 = 36.

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sandeepd: 2020-01-27 10:13:54

Cool problem indeed!

Sigma Kappa: 2017-09-18 19:28:42

I wish current GCJ's 1st round problems were like this... BTW this is a carbon copy of http://www.spoj.com/problems/BRKSTRNG/

kshubham02: 2016-06-04 13:27:06

Be careful about the output format. I was displaying just the numbers and it costed me two WA.

SUBHAM SANGHAI: 2016-05-25 08:05:26

Nice dp .. my 100 th :D

maverick_10: 2016-02-03 14:34:44

A very good dp problems.
Do solve on your own.

naruto09: 2016-01-23 07:20:06

beautiful question....:) :D

@DubeY@: 2015-08-06 20:33:22

please provide any spacial case......getting back to back WA !!

kelaseek: 2015-01-08 05:32:55

you should memoize

Tapashee Tabassum Urmi: 2014-12-29 17:58:31

Awesome solution for large dataset. :)

a1160576: 2014-07-18 12:35:13

I was equal fastest with least memory amongst solutions that speed, can't complain.

Was a good problem that I tried entirely the wrong approach for at first.

Last edit: 2014-07-18 12:35:40

Added by:Shafaet
Time limit:20s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64
Resource:Google Codejam 2009, Round 1C, Problem C