# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3821 |
3 | Benq | 3736 |
4 | Radewoosh | 3631 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3388 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
Name |
---|
I'll try to explain my solution. Firstly, I hope you get that you never need to make more than 2 triples of the format [x, x + 1, x + 2] (Claim 1). Now create a tile frequency array from 1 to m. We'll apply DP on this starting from 1 till m. DP state is dp[i][j][k], where i is the last index of the prefix considered, j is the number of tiles remaining of size (i — 1) and k is the number of tiles remaining of size i. Because of the claim 1, we never need to store more than 2 tiles of size (i — 1) and 4 tiles of size i for later operation (Why?). To make the transition to (i + 1), you can iterate for all possible values of j and k. https://codeforces.me/contest/1110/submission/49606713. (Note that DP state with -1 as value is representing the impossible state.)
Firstly, I hope you get that you never need to make more than 2 triples of the format [x, x + 1, x + 2] --> sorry , but i didn't get . what will happen if we take 3 triplets , will it lead to wa ?
No. Because you can always replace 3 triples of this format to 3 triples [x, x, x], [x + 1, x + 1, x + 1] and [x + 2, x + 2, x + 2].
At first, let's notice that it isn't worth taking 3 times or more [x, x + 1, x + 2], we can take [x, x, x], [x + 1, x + 1, x + 1], [x + 2, x + 2, x + 2] instead.
Now let's compute how many times number x is present for any valid x. Denote the value by cnt[x].
Now let's calculate dp[i][j][k] (in this dp we consider all numbers from the given array that are less than i - 1, cnt[i - 1] - k (k is up to 3) numbers which are equals to i - 1 and j numbers which are equal to i (j is up to cnt[i]), dp[i][j][k] is the answer for these numbers)
How to calculate it? As we know, there is no point in taking 3 times [x, x + 1, x + 2], so dp[i][j][k] = max(dp[i - 1][cnt[i] - l][l] + (j - l) / 3), l < 3. That's because we can take [i - 2, i - 1, i] 0, 1 or 2 times. If we take it l times, than we have (j - l) numbers that are equals to i and we have to divide them into groups of the type [i, i, i]. Also we have to divide the other numbers, but we know that the answer, it is dp[i — 1][cnt[i] — l][l]
Hope, my answer will help you. Ask me if you have any questions :D
thanks i get it . Miraak