In a weighted tree, how to find for some node (u) the distance to another node (v) (answering Q queries effeiciently)? Constraints N <=10^3, Queries <=10^3
# | User | Rating |
---|---|---|
1 | tourist | 3856 |
2 | jiangly | 3747 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3591 |
6 | gamegame | 3477 |
7 | Benq | 3468 |
8 | Radewoosh | 3462 |
9 | ecnerwala | 3451 |
10 | heuristica | 3431 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | -is-this-fft- | 162 |
3 | Dominater069 | 160 |
4 | Um_nik | 158 |
5 | atcoder_official | 156 |
6 | Qingyu | 152 |
6 | djm03178 | 152 |
6 | adamant | 152 |
9 | luogu_official | 149 |
10 | awoo | 147 |
In a weighted tree, how to find for some node (u) the distance to another node (v) (answering Q queries effeiciently)? Constraints N <=10^3, Queries <=10^3
Name |
---|
Since N is small you can apply BFS starting from u in each query and print the distance of node v which will take O(n) time for each query. An efficient way to answer each query in O(logn) is using LCA which can be found in O(logn) using binary lifting.
I teach all these concepts with practice problems , here is the graph theory playlist where you can find this concept along with other concepts. https://www.youtube.com/playlist?list=PL2q4fbVm1Ik64I3VqbVGRfl_OgYzvzt9m hope this helps.