№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3839 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3612 |
7 | Geothermal | 3569 |
8 | ecnerwala | 3494 |
9 | Um_nik | 3396 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | Um_nik | 164 |
2 | -is-this-fft- | 162 |
3 | maomao90 | 159 |
3 | atcoder_official | 159 |
5 | cry | 158 |
5 | awoo | 158 |
7 | adamant | 155 |
8 | nor | 154 |
9 | TheScrasse | 153 |
10 | Dominater069 | 152 |
Название |
---|
Petr For the TopCoder problem, mentioned, I'm unable to prove this part:
However, we still have the freedom of choosing which pair of green and red ends we use for reducing the problem to size n-1. If b>0, then we will choose which green end is one of the red ends of the first green-red string paired with.
If we delay merging green-green strings with red-red strings until the end, how do we prove that the answer doesn't change? Playing around with the DP recurrence didn't help.
Thanks for your help!