WALL-E__'s blog

By WALL-E__, history, 3 months ago, In English,

i am having problem in solving this problem using 0/1 knapsack problem. http://codeforces.com/contest/19/problem/B

 
 
 
 
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3 months ago, # |
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You need to keep two states:

I: Current item.

R: Number of items remaining.

The transition is, take the item I paying ci and go to the item I + 1 with R - 1 - ti remaining items, or dont take the item I and go to the item I + 1 with R remaining items.

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3 months ago, # |
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There's a simple solution that runs in O(2N). Just try all combinations and keep the best.

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    3 months ago, # ^ |
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    But 1 ≤ N ≤ 2000, do you have some super computer ?

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      3 months ago, # ^ |
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      Well, not me, but the studios where I usually work make 3D movies too, so they have a farm of around 45 powerful computers running at the same time.

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        3 months ago, # ^ |
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        Wow, in how many seconds them run 22000 operations ?, can you give me one powerful computer ?

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          3 months ago, # ^ |
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          Well, you can do the math. One powerful computer can do 1000 million operations per second, which is around 230, so it can do 60 * 230 in a minute. That's around 236 in a minute and 242 in one hour, so it would take something like 21958 hours to run the solution. One year has 213 hours approximately, so it will take 21945 years. Is it good enough?

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            3 months ago, # ^ |
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            Complexity if exponential, but you can get it to run under 1000ms (in java) using the sleep.thread() trick. This should be common knowlegde.

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              3 months ago, # ^ |
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              Indeed, or you can use #pragma lightspeed in C++55, which will be released in 37 years. But since it uses black hole quantum physics to compile and run, it doesn't matter, you can use it today by calling that pragma.

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                3 months ago, # ^ |
                Rev. 2   Vote: I like it +6 Vote: I do not like it

                So it turns out C++ really is better at everything. I think using a bitset will reach O(1/n) (watch out for n = 0) complexity, but seems that its too difficult for a simple problem as this

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                  3 months ago, # ^ |
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                  well they didn't make minecraft in c++ did they

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3 months ago, # |
Rev. 2   Vote: I like it -6 Vote: I do not like it

Another way is [itemupto][time] with observation time > 2000 dont matter. do we store dp[item][time] = minimum cost.

Note that for each item we can "add" 1 second to it because it takes care of itself.

And we us the dp table to solve problem.

EDIT: Wow it seems I've greatly infected the integrity of CF by giving out correct solutions! The horror!