My first blog post :) Any constructive feedback is welcome.
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Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
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I think there is a O(N * M * 2N) solution for this problem. DP[i][j][p] = solution where we currently start at point (i, j) and the first i-bits of p belongs to the (j + 1)-column and the rest of the bits belongs to the j-column. And transitions are similar to the last solution. This utilizes the 2N different values of p and q more efficiently (since in the other solution, p cannot have zero bit before the last set bit, and q cannot have set bit after the row we currently stand in).
This is what I think is the actual idea behind broken profile dp.
can you provide a link to the iterative code for the same?
https://vjudge.net/problem/22385/origin
how to solve this
Thanks you very much
Is the blog still available? I have trouble with your link.
Sorry about that. Updated the link :)
Thanks:)
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Auto comment: topic has been updated by evilbuggy (previous revision, new revision, compare).