[algogeeks] Re: find a way

2011-01-26 Thread ritu
@above
How can you calculate dp[n][0] with above recursive eq??


On Jan 23, 5:40 pm, sankalp srivastava richi.sankalp1...@gmail.com
wrote:
 @ above
 In that case  , it will be a simple dynamic programming based
 recursion

 assuming the total distance one has to cover is n ;
 d[i][j]=minimum number of fuel stations to stop at in order to cross i
 stations and with j miles still to go .
 dp[n][0]= minimum number of fuel stations to stop at in order cross n
 stations and with 0 miles still to go (Assuming the nth stop coincides
 with the destination B .In case it does not , we can answer something
 like dp[n][p]  , where p is the distance to go from nth stop to A)

 The recursion
 dp[i][k]= min(dp[i+1][k- distance b/w the ith and (i+1)th fuel
 station] , dp[i+1][k- distance +lk]+1)(lk= distance we can cover on
 this stop)

 base case dp[0][j]=0;(for each j )// we have to cover no more stations
 therefore

 On Jan 22, 9:40 pm, Divya Jain sweetdivya@gmail.com wrote:

  if u can take only a certain amount of fuel from a particaular station ie
  station xi can provide li amoutn of fuel.. then wat?

  On 22 January 2011 13:46, Terence technic@gmail.com wrote:

   Greedy-Approach.
   Refueling only when you have to.

   On 2011-1-22 15:59, snehal jain wrote:

   Suppose you want to travel from city A to city B by car, following a
   fixed route. Your car can travel m miles on a full tank of gas, and
   you have a map of the n gas stations on the route between A and B that
   gives the number of miles between each station.
   Design an algorithm to find a way to get from A to B without running
   out of gas that minimizes the total number of refueling stops, in O(n)
   time.

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[algogeeks] Re: find a way

2011-01-23 Thread sankalp srivastava
@ above
In that case  , it will be a simple dynamic programming based
recursion

assuming the total distance one has to cover is n ;
d[i][j]=minimum number of fuel stations to stop at in order to cross i
stations and with j miles still to go .
dp[n][0]= minimum number of fuel stations to stop at in order cross n
stations and with 0 miles still to go (Assuming the nth stop coincides
with the destination B .In case it does not , we can answer something
like dp[n][p]  , where p is the distance to go from nth stop to A)

The recursion
dp[i][k]= min(dp[i+1][k- distance b/w the ith and (i+1)th fuel
station] , dp[i+1][k- distance +lk]+1)(lk= distance we can cover on
this stop)

base case dp[0][j]=0;(for each j )// we have to cover no more stations
therefore

On Jan 22, 9:40 pm, Divya Jain sweetdivya@gmail.com wrote:
 if u can take only a certain amount of fuel from a particaular station ie
 station xi can provide li amoutn of fuel.. then wat?

 On 22 January 2011 13:46, Terence technic@gmail.com wrote:



  Greedy-Approach.
  Refueling only when you have to.

  On 2011-1-22 15:59, snehal jain wrote:

  Suppose you want to travel from city A to city B by car, following a
  fixed route. Your car can travel m miles on a full tank of gas, and
  you have a map of the n gas stations on the route between A and B that
  gives the number of miles between each station.
  Design an algorithm to find a way to get from A to B without running
  out of gas that minimizes the total number of refueling stops, in O(n)
  time.

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  You received this message because you are subscribed to the Google Groups
  Algorithm Geeks group.
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  .
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 http://groups.google.com/group/algogeeks?hl=en.

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