问题
I use a matrix d
to present a graph. d.(i).(j)
means the distance between i
and j
; v
denotes the number of nodes in the graph.
It is possible that there is negative cycle in this graph.
I would like to check if a negative cycle exists. I have written something as follows from a variation of Floyd-Warshall:
let dr = Matrix.copy d in
(* part 1 *)
for i = 0 to v - 1 do
dr.(i).(i) <- 0
done;
(* part 2 *)
try
for k = 0 to v - 1 do
for i = 0 to v - 1 do
for j = 0 to v - 1 do
let improvement = dr.(i).(k) + dr.(k).(j) in
if improvement < dr.(i).(j) then (
if (i <> j) then
dr.(i).(j) <- improvement
else if improvement < 0 then
raise BreakLoop )
done
done
done;
false
with
BreakLoop -> true
My questions are
- Is the code above correct?
- Is the
part 1
useful?
Because I call this function very often, I really want to make it as fast as possible. So my 3) question is if other algorithms (especially Bellman-Ford
) can be quicker than that?
回答1:
The first question about the correctness of your code is more appropriate for http://codereview.stackexchange.com.
Either of Bellman-Ford or Floyd-Warshall is appropriate for this problem. A comparison of performance follows:
- Bellman-Ford (Wikipedia)
- Time-complexity:
O(|V|*|E|)
- Space-complexity:
O(|V|)
- The section on "Finding negative cycles" is what you want
- Time-complexity:
- Floyd-Warshall (Wikipedia)
- Time-complexity:
O(|V|^3)
- Space-complexity:
O(|V|^2)
- Time-complexity:
Since |E|
is bounded by |V|^2
, Bellman-Ford is the clear winner and is what I would advise you use.
If graphs without negative cycles is the expected normal case, it might be appropriate to do a quick check as the first step of your algorithm: does the graph contain any negative edges? If not, then it of course does not contain any negative cycles, and you have a O(|E|)
best case algorithm for detecting the presence of any negative cycles.
来源:https://stackoverflow.com/questions/16898416/fastest-algorithm-to-detect-if-there-is-negative-cycle-in-a-graph