I have built a brute force solution to this problem but of course time limit exceeded monster has appeared!
please can anyone explain a solution or just give a hint for this problem, i have read the editorial but i didn't understand what is written.
# | User | Rating |
---|---|---|
1 | jiangly | 3898 |
2 | tourist | 3840 |
3 | orzdevinwang | 3706 |
4 | ksun48 | 3691 |
5 | jqdai0815 | 3682 |
6 | ecnerwala | 3525 |
7 | gamegame | 3477 |
8 | Benq | 3468 |
9 | Ormlis | 3381 |
10 | maroonrk | 3379 |
# | User | Contrib. |
---|---|---|
1 | cry | 168 |
2 | -is-this-fft- | 165 |
3 | Dominater069 | 161 |
4 | Um_nik | 159 |
4 | atcoder_official | 159 |
6 | djm03178 | 157 |
7 | adamant | 153 |
8 | luogu_official | 150 |
9 | awoo | 149 |
10 | TheScrasse | 146 |
I have built a brute force solution to this problem but of course time limit exceeded monster has appeared!
please can anyone explain a solution or just give a hint for this problem, i have read the editorial but i didn't understand what is written.
recently i have been studying graph theory from competitive programming book ,in applications of DFS there was articulation points and bridges . He defines other than visited array a another one here is the statement.
[This algorithm maintains two numbers: dfs_num(u) and dfs_low(u). Here, dfs_num(u) stores the iteration counter when the vertex u is visited for the first time and not just for distinguishing DFS_WHITE versus DFS_GRAY/DFS_BLACK. The other number dfs_low(u) stores the lowest dfs_num reachable from DFS spanning sub tree of u. Initially dfs_low(u) = dfs_num(u) when vertex u is first visited. Then, dfs_low(u) can only be made smaller if there is a cycle (some back edges exist). Note that we do not update dfs_low(u) with back edge (u, v) if v is a direct parent of u. ]
please anyone give an explanation for this part.
Name |
---|