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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 :-)