| # | User | Rating |
|---|---|---|
| 1 | jiangly | 3810 |
| 2 | Benq | 3676 |
| 3 | Kevin114514 | 3655 |
| 4 | maroonrk | 3463 |
| 5 | strapple | 3447 |
| 6 | Um_nik | 3387 |
| 7 | heuristica | 3322 |
| 8 | turmax | 3317 |
| 9 | tourist | 3307 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | Um_nik | 141 |
| 5 | Errichto | 139 |
| 6 | adamant | 137 |
| 7 | AmShZ | 135 |
| 8 | BledDest | 132 |
| 9 | maroonrk | 131 |
| 10 | qwexd | 129 |
| Name |
|---|



Let Level(X) will be index of vertice X in some topologically sorted vector of vertices. DP(x1, x2, x3) — maximum possible length of 3-path (x1, x2, x3) -> (t1, t2, t3). to calculate DP(x1, x2, x3) you shuld try to move from some of three vertices (x1, x2, x3) with minimum Level.
UPD: vertices are already topologically sorted, Level(x) == x.
It's O(N^4). Maybe there is better solution.
thank you very much
but is what is wrong with my solution ?!
It's wrong. Even bfs is not a BFS.
is that because it can be longest 3-crtical path that the 3 paths are not the longest for each path ??
I did't get your question but I think answer is "Yes".
your bfs does'nt finds longest path.
that's what i meant... thank you again :-)