Hi, everyone.
Can someone, please, explain the solution for this problem ?
I really cannot understand what editorial wants to say.# | User | Rating |
---|---|---|
1 | jiangly | 4039 |
2 | tourist | 3841 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3590 |
5 | ecnerwala | 3542 |
6 | Benq | 3535 |
7 | orzdevinwang | 3526 |
8 | gamegame | 3477 |
9 | heuristica | 3357 |
10 | Radewoosh | 3355 |
# | User | Contrib. |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | atcoder_official | 160 |
3 | Um_nik | 160 |
5 | djm03178 | 158 |
6 | Dominater069 | 156 |
7 | adamant | 153 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 147 |
Hi, everyone.
Can someone, please, explain the solution for this problem ?
I really cannot understand what editorial wants to say.Name |
---|
Auto comment: topic has been translated by TigranHakobyan(original revision, translated revision, compare)
Sorry that I can't find editorial. I think you can sort all cars alphabetically and use DP to find maximum length. You have to sort cars to find first-alphabetical one easily.
If you mean just sort a alphabetically, then if we consider {acbd, abcd, bda}. Using your idea we will get answer 2 but the answer is 3, isn't it ?
P.S. editorial
Probelm says, "The front end is the end with the letter that comes first in the alphabet", so you have to flip car "bda" into "adb".
so in case {"acbd", "abcd", "bda"}, the answer is 1.
and if the car A and car B can be linked, A.front <= A.back == B.front <= B.back, so sort like this.
ardiankp's code
sorry about my English — I'm not good at it.
Thank you, so much. I misunderstood the problem.