I don't understand, why this simple problem, solved only 25 coders!

Re: I don't understand, why this simple problem, solved only 25 coders!

Posted by

svr 14 Feb 2007 22:46

I will try tomorrow and will be in one of two parties

Re: I don't understand, why this simple problem, solved only 25 coders!

Posted by

Kit 15 Feb 2007 01:16

As for me, this problem has quite unusual idea. BTW, I heard what famous "hard life" has fairly simple solution :)

Re: I don't understand, why this simple problem, solved only 25 coders!

May be you one of the best coders:) an it's simple for you?

The solution is really easy, but it is not that easy to think it out (-)

Re: The solution is really easy, but it is not that easy to think it out (-)

Posted by

svr 16 Feb 2007 00:08

Could'n found very simple solution.

In final version calcul z1(k)= min [F(0,0,a1)+F(0,a1,a2)+

F(a1,a2,a3)+...+F(ak-1,ak,0)+F(ak,0,0)- a1-...-ak]

for all [a1....ak] k=1,....

z2(k)=min [-F(0,0,a1)-F(0,a1,a2)+

-F(a1,a2,a3)+...-F(ak-1,ak,0)+F(ak,0,0)+ a1-...+ak]

if (z1<0)&&(z2<0) <> and so on.

k from 1 to 10000 but functional depend only ak-1,ak

and z1(k-2). History fogotten this is "fishka" in my prog

Ac(1.3 cek) Ho make 0.015 don't now .

Re: The solution is really easy, but it is not that easy to think it out (-)

AC(0.109) using DP, but i was really surprised, when my program got AC

Re: The solution is really easy, but it is not that easy to think it out (-)

dp with bfs - i think it is very standart algo

Re: The solution is really easy, but it is not that easy to think it out (-)

Posted by

svr 8 Mar 2007 17:24

My method also standart.

It is Bellman-method for shortest path when

functional of length rather specific but have

main property of fogotten history

Re: The solution is really easy, but it is not that easy to think it out (-)

I think, this method is DP

Re: The solution is really easy, but it is not that easy to think it out (-)

It's quite obvious DP. Search for positive/negative loop around (0,0) in graph with transfers (i,j) -> (j,k)