Hi, everyone.
Can someone, please, explain the solution for this problem ?
I really cannot understand what editorial wants to say.# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
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.