检测图形中是否存在负周期的最快算法

检测图形中是否存在负周期的最快算法

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问题描述

我使用矩阵 d 表示图表。 d。(i)。(j)表示 i 和 j之间的距离; v 表示图中的节点数。



此图中可能存在负循环。



我想检查是否存在负循环。我从的变体中写了以下内容:

  let dr = Matrix.copy d in 

(* part 1 *)
for i = 0至v-1做
博士(i)。(i)完成;

(*第2部分*)
尝试
for k = 0到v-1做
for i = 0到v-1做
j = 0至v-1做
让改善=
中的(i)。(k)+(k)博士。博士(i)。(j)然后(
如果(i< j))然后
博士(i)。(j)<-改善
如果改善< ; 0然后
提高BreakLoop)
完成
完成
完成;
错误

BreakLoop->真

我的问题是


  1. 上面的代码正确吗?

  2. 第1部分有用吗?

因为我经常调用此函数,所以我真的想使其尽可能快。所以我的3)问题是,其他算法(尤其是 Bellman-Ford )是否可以比这更快?

解决方案

关于代码正确性的第一个问题更适合。






或适用于此问题。性能比较如下:




  • (维基百科)

    • 时间复杂度: O(| V | * | E |)

    • 空间复杂度: O(| V |)

    • 是您想要的


  • (维基百科)

    • 时间复杂度: O(| V | ^ 3)

    • 空间复杂度: O(| V | ^ 2)




因为 | E | 受限制 | V | ^ 2 ,是最重要的ar赢家,我会建议您使用它。






如果图没有的负周期是在预期的正常情况下,作为算法的第一步,可能需要进行快速检查:图形是否包含负边缘?如果不是,那么它当然不包含任何负循环,并且您有一个 O(| E |)最佳情况算法来检测是否存在任何负循环。 / p>

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

  1. Is the code above correct?
  2. 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?

解决方案

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:

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.

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08-16 07:20