My first blog post :) Any constructive feedback is welcome.
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 166 |
2 | maomao90 | 163 |
2 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | adamant | 160 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | nor | 153 |
9 | Dominater069 | 153 |
My first blog post :) Any constructive feedback is welcome.
Name |
---|
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:)
Auto comment: topic has been updated by evilbuggy (previous revision, new revision, compare).
Auto comment: topic has been updated by evilbuggy (previous revision, new revision, compare).
Auto comment: topic has been updated by evilbuggy (previous revision, new revision, compare).