Блог пользователя prabhat7298

Автор prabhat7298, история, 5 лет назад, По-английски

In most of the knapsack variants we've a linear dependency on the knapsack size M but in case where we've many small items leading to a very large knapsack capacity we need an alternate way of solving it. I read about it here that we can solve it by using shortest path algorithm but I wasn't able to grab the whole concept. Can anyone explain it in simpler words and can comment a little on the implementation part. Here you can find a related problem.

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Interesting post!

Can anyone explain why the complexity stated in the first link is $$$O(w^\frac{3}{2} + nw)$$$ and not $$$O(w^2 + nw)$$$? Also, why does the solution of complexity $$$O(n w \log{w})$$$ pass the tests of the problem in no time? Should that not exceed TL?

Solution

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    5 лет назад, # ^ |
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    bicsi Thanks for the code. understood it! Btw is it possible to answer queries asking maximum value we can obtain for a weight if along with original weights corresponding values are also given?