ANARC05B - The Double HeLiX


Two finite, strictly increasing, integer sequences are given. Any common integer between the two sequences constitute an intersection point. Take for example the following two sequences where intersection points are
printed in bold:

  • First= 3 5 7 9 20 25 30 40 55 56 57 60 62
  • Second= 1 4 7 11 14 25 44 47 55 57 100

You can ‘walk” over these two sequences in the following way:

  1. You may start at the beginning of any of the two sequences. Now start moving forward.
  2. At each intersection point, you have the choice of either continuing with the same sequence you’re currently on, or switching to the other sequence.

The objective is finding a path that produces the maximum sum of data you walked over. In the above example, the largest possible sum is 450, which is the result of adding 3, 5, 7, 9, 20, 25, 44, 47, 55, 56, 57, 60, and 62

Input

Your program will be tested on a number of test cases. Each test case will be specified on two separate lines. Each line denotes a sequence and is specified using the following format:

n v1 v2 ... vn

Where n is the length of the sequence and vi is the ith element in that sequence. Each sequence will have at least one element but no more than 10,000. All elements are between -10,000 and 10,000 (inclusive).
The last line of the input includes a single zero, which is not part of the test cases.

Output

For each test case, write on a separate line, the largest possible sum that can be produced.

Sample

Input:
13 3 5 7 9 20 25 30 40 55 56 57 60 62
11 1 4 7 11 14 25 44 47 55 57 100
4 -5 100 1000 1005
3 -12 1000 1001
0

Output:
450
2100

hide comments
sonali9696: 2016-12-20 09:47:12

Damn! Don't overthink like me. It will cost you too much time. Think like a kid. Very direct

pratyush2311: 2016-12-12 08:59:38

No Dp. Just simple logic.Think for a while about how to proceed and just implement the logic :)

surf_excel: 2016-12-05 11:42:58

some testcases are separated by uneven number of spaces. Using split() in java landed me in NZEC couple of times .... changing it to split(" +") made it accepted :)

jawad_cs: 2016-11-29 10:34:06

guyz don't waste ur time on trying to figure out a DP solution it can be solved in O(n) with simple logic

caphindsight: 2016-10-25 09:24:14

Btw note to those who get WAs because of the boundary cases. Just append (intersection point) z = max(a.back(), b.back()) + 1 to the end of both sequences, then subtract z from your answer. This way no special code is required for dealing with boundary cases.

Last edit: 2016-10-25 09:24:45
caphindsight: 2016-10-25 09:17:08

I was looking for a DP problem to solve... Stumbled upon this. While it is an OK beginner problem, I hate the tags... No DP required! No binary search required! This problem can be solved by two pointers loop, with O(n) asymptotic complexity.

Last edit: 2016-10-25 09:18:21
rraj001: 2016-10-13 06:22:06

dp+binary serach ! :)

siddharth_0196: 2016-10-11 17:06:14

Simple logic will work! No DP!

Last edit: 2016-10-11 18:58:31
davidgalehouse: 2016-10-06 04:12:13

Note to C# solvers: input is formatted incorrectly, you should RemoveEmptyEntries if splitting.

Last edit: 2016-10-06 04:12:27
Anonomous: 2016-10-02 15:53:23

Be careful, some of the test cases are not strictly increasing.


Added by:~!(*(@*!@^&
Date:2009-07-05
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ERL JS-RHINO NODEJS PERL6 VB.NET
Resource:ANARC 2005