@ 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 <[email protected]> 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 <[email protected]> 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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