我用的矩阵 D 来present图。 D。（我）。（J）指的的距离我和Ĵ; v 表示在图中节点的数目。

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.

有可能存在在此图中的负循环。

It is possible that there is negative circle in this graph.

我想检查是否负圈存在。我已经写了一些东西弗洛伊德 - 沃肖尔的变化如下：

I would like to check if a negative circle 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

我的问题是

是code以上正确吗？ 是第1部分有用吗？ Is the code above correct? Is the part 1 useful?

由于我调用该函数很多时候，我真的想让它尽可能快。所以，我的3个）问题是，如果其他算法（尤其是贝尔曼 - 福特）可以比快？

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?

推荐答案

有关您code是正确的第一个问题是更适合的 HTTP：//$c$creview.stackexchange.com

The first question about the correctness of your code is more appropriate for http://codereview.stackexchange.com.

无论贝尔曼 - 福特或的弗洛伊德 - 沃肖尔是适合这个问题。业绩比较如下：

Either of Bellman-Ford or Floyd-Warshall is appropriate for this problem. A comparison of performance follows:

贝尔曼 - 福特（维基百科） 时间复杂度： O（| V | * | E |） 空间复杂度： O（| V |） 的查找负循环的部分是什么你想 Bellman-Ford (Wikipedia) Time-complexity: O(|V|*|E|) Space-complexity: O(|V|) The section on "Finding negative cycles" is what you want 时间复杂度： O（| V | ^ 3） 空间复杂度： O（| V | ^ 2） Time-complexity: O(|V|^3) Space-complexity: O(|V|^2)

由于 | E | 是界| V | ^ 2 ，贝尔曼 - 福特是明显的赢家，是什么，我会建议你使用。

Since |E| is bounded by |V|^2, Bellman-Ford is the clear winner and is what I would advise you use.

如果图的没有的负循环是预期的正常情况下，它可能是适当的做一个快速检查作为你的算法的第一步：请问图中包含任何负面的边缘？如果没有，那么它当然不包含任何负面的周期，和你有一个 O（| E |）最好的情况下检测算法任何负面的presence周期。

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.