№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3831 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | gamegame | 3386 |
10 | ksun48 | 3373 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
Название |
---|
Can anybody from community can help with me this question ? Its been 7hrs and I dont know why people doesnt respond to question that may look small to you but its like mount Everest for me. :/
dp[c][i] — minimal cost of painting the first i columns in NxM rectangle into a barcode the way its last stripe will be (c == 0 ? black : white).
This line checks if it is optimal to paint last a lines in black.
And this line checks if it is optimal to paint last a lines in white.
Can you please explain how the two recurrence relation is formed ?
Well, let's look on Nxi rectangle formed with the first i columns. What can we say about its last stripe? Either it's black or white and his width is ranged from x to y. Therefore, we handle 2 * (y — x + 1) variants. Let's imagine the last stripe in your subrectangle is white. It means that previous stripe's color has to be the different (black). Then, for each a x <= a <= y check the cost of oainting rectangle Nx(i-a) in stripes so it will end with the black stripe (dp[0][i-a])and add the cost of painting columns from i-a+1 to i in white (sum-of-whites(i-a+1,i)). Then we keep the best outcome.
Same thing with the case, where the last stripe in rectangle Nxi is white, but with reverse colouring.