| № | Пользователь | Рейтинг |
|---|---|---|
| 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 |
| 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 | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
A tree is given in which each edge is having some weight associated with it and you are given a number K.
So starting from root in K-steps how much maximum weight you can collect. You can traverse in any direction either from parent to child or child to parent. You can visit a node multiple times.
~~~~~ ~~~~~
I have not formally proved this, but say we use the edges e1, e2, e3, ..., ek (these are the weights). So sum over all ei is less than k * max(ei), i.e, it is better to repeat the max edge k times. So this may lead to a greedy algorithm. For each edge, if the depth of the node is d, add the weight upto that depth, and then add the edge k - d times. Find the maximum for all edges.