My first blog post :) Any constructive feedback is welcome.
| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
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).