There is no algorithm which can find all the cycles in a directed graph in polynomial time. Suppose, the directed graph has n nodes and every pair of the nodes has connections to each other which means you have a complete graph. So any non-empty subset of these n nodes indicates a cycle and there are 2^n-1 number of such subsets. So no polynomial time algorithm exists. So suppose you have an efficient (non-stupid) algorithm which can tell you the number of directed cycles in a graph, you can first find the strong connected components, then applying your algorithm on these connected components. Since cycles only exist within the components and not between them.

There is no algorithm which can find all the cycles in a directed graph in polynomial time. Suppose, the directed graph has n nodes and every pair of the nodes has connections to each other which means you have a complete graph. So any non-empty subset of these n nodes indicates a cycle and there are 2^n-1 number of such subsets. So no polynomial time algorithm exists. So suppose you have an efficient (non-stupid) algorithm which can tell you the number of directed cycles in a graph, you can first find the strong connected components, then applying your algorithm on these connected components. Since cycles only exist within the components and not between them.

我已经在sml(命令式编程)中实现了这个问题。这是大纲。找到所有入度或出度为0的节点。这样的节点不能成为循环的一部分(因此将它们删除)。接下来，从这些节点中删除所有传入或传出边。 递归地将此过程应用于结果图。如果最后你没有剩下任何节点或边，图就没有任何循环，否则就有。

从DFS开始:当且仅当DFS期间发现后边缘时，循环存在。这是白径定理的结果。

假设这是一个作业时间表，我怀疑在某些时候您会将它们按照建议的执行顺序进行排序。

如果是这种情况，那么拓扑排序实现在任何情况下都可以检测到循环。UNIX tsort当然可以。因此，我认为在三步排序的同时检测循环比在单独的步骤中检测更有效。

因此，问题可能变成“我如何最有效地进行tsort”，而不是“我如何最有效地检测循环”。答案可能是“使用图书馆”，但如果没有下面的维基百科文章:

http://en.wikipedia.org/wiki/Topological_sorting

有一种算法的伪代码，以及来自Tarjan的另一种算法的简要描述。两者都具有O(|V| + |E|)时间复杂度。

如果DFS发现一条边指向一个已经访问过的顶点，那么这里就有一个循环。