Given a string Str, rearrange Str such that the resultant string T maximizes min (LCS(Str, T) and LCS(Str, reverse(T))).
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 155 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | djm03178 | 152 |
Given a string Str, rearrange Str such that the resultant string T maximizes min (LCS(Str, T) and LCS(Str, reverse(T))).
Название |
---|
looks like min (LCS(Str, T) and LCS(Str, reverse(T))) cannot exceed longest palindromic subsequence of Str, hence T=Str should work, i may be wrong tho
Actually it can exceed that.
Let $$$str = \text{aabb}$$$, $$$T = \text{abba}$$$. Now, $$$\min(\mathrm{LCS}(\text{aabb}, \text{abba}), \mathrm{LCS}(\text{aabb}, \text{abba})) = \min(3, 3) = 3$$$.
If you choose $$$T = str$$$, you get $$$\min(\mathrm{LCS}(\text{aabb}, \text{aabb}), \mathrm{LCS}(\text{aabb}, \text{bbaa})) = \min(4, 2) = 2$$$.
yeah, i knew most probably my claim must be wrong
anyways, please tell how to solve the above problem