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.
~~~~~ ~~~~~
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3690 |
| 2 | jiangly | 3647 |
| 3 | Benq | 3581 |
| 4 | orzdevinwang | 3570 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | Radewoosh | 3509 |
| 8 | ecnerwala | 3486 |
| 9 | jqdai0815 | 3474 |
| 10 | gyh20 | 3447 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 174 |
| 2 | awoo | 164 |
| 3 | adamant | 163 |
| 4 | TheScrasse | 159 |
| 5 | nor | 157 |
| 6 | maroonrk | 156 |
| 7 | -is-this-fft- | 152 |
| 8 | Petr | 146 |
| 8 | orz | 146 |
| 10 | BledDest | 145 |
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.