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 | jiangly | 3976 |
2 | tourist | 3815 |
3 | jqdai0815 | 3682 |
4 | ksun48 | 3614 |
5 | orzdevinwang | 3526 |
6 | ecnerwala | 3514 |
7 | Benq | 3482 |
8 | hos.lyric | 3382 |
9 | gamegame | 3374 |
10 | heuristica | 3357 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | -is-this-fft- | 165 |
3 | Um_nik | 161 |
3 | atcoder_official | 161 |
5 | djm03178 | 157 |
6 | Dominater069 | 156 |
7 | adamant | 154 |
8 | luogu_official | 152 |
9 | awoo | 151 |
10 | TheScrasse | 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.