At time t=10.01, the cars are at positions: A - [40.04,45.04] B - [49.02,54.02] C - [45.01,50.01] D - [54.01,59.01]
Note that the front of A is next to the back of C. They must be in different lanes, and must stay in those lanes until A fully passed C at time 13.3333. Note also that the front of C is next to the back of A. C and A must therefore be in different lanes, and so A and B must be in the same lane. Note also that the front of B is next to the back of D. They must be in different lanes, and must stay in those lanes until B fully passes D at time 20. So to summarise, from time 10.01 to time 13.333, cars A and B are in the same lane and cannot change lane. Therefore, at time 12 when A reaches B, maximal bang occurs. On 29 Apr 2012, at 21:16, xenoky <[email protected]> wrote: > Hi, > > The 4th test case in the Cruise Control problem has the following > input: > 4 > L 4 0 > L 2 29 > L 1 35 > L 1 44 > > The given solution is: > Case #4: 12 > > calculating the distance at t=12 > > 4*12=48 > 2*12+29=53 > 12+35=47 > 12+44=56 > > ordering this distance [47,48,53,56] > > The maximum distance beetween nearest triplet is 6! -> [47,48,53] > > I can put the car this way > > ##### ##### > ##### > > without accident > > Where am i wrong? > > My program answer is 13... > > -- > You received this message because you are subscribed to the Google Groups > "Google Code Jam" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/google-code?hl=en. > -- You received this message because you are subscribed to the Google Groups "Google Code Jam" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/google-code?hl=en.
