I have a idea for this problem, but I cannot
ensure whether the solution is correct.
At First, I guess that What's solution to get
is that to choice the root of the deepest subtree
in the relationship in each superior officer to
direct subordinate.
( If the guessing is correct, then I have a suggestion
as following, And if the guessing is wrong,
you couldn't waste time to see the following )
What's important for the problem to focus on is that
if we can figure out every superior officer' s longest
deep into the leave, then the problem will solve since
it could decide the order of rounds for each
superior officer.
We may not to figure otu the longest deep,
if we can decide the order of rounds for each
superior officer before.
Now, I have a method to
decide the order of rounds for each superior office
as the following...
----------------------------------------------------------------------------
Suppose T is the tree about the problem that to
notify postpone in fast way from superior officer
to direct subordinate by phone calling.
V(T) = {v|v is node of T}
E(T) = {(v,u)| v is superior officer of the subordinate u}
leaves(T) := {v in V(T)| v is leave of T}
construct graph H s.t.V(H)={v|v in V(T)}∪{dummy} and
E(H)={(v,u)|(u,v) in E(T)}∪{(dummy,v)| leaf in v(T) }
I presume that every body know BFS algorithm.
By
using breath searching that dummy initialize to search
graph H with a quene Q but allowing
each vertex in H can be repeatly visited to distinguish
BFS Algorithm,
we know that
the order of round for each vertex v in V(T)
is reverse order of visit for v in V(H).
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