正如你所说，你有一组作业，它需要按一定的顺序执行。给定作业调度所需顺序的拓扑排序(如果是直接的非循环图，则用于解决依赖问题)。运行dfs并维护一个列表，并在列表的开头开始添加node，如果您遇到一个已经被访问过的节点。然后在给定的图中找到一个循环。

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。

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.

最简单的方法是对图进行深度优先遍历(DFT)。

如果图有n个顶点，这是一个O(n)时间复杂度算法。因为你可能必须从每个顶点开始进行DFT，所以总复杂度变成O(n^2)。

您必须维护一个包含当前深度第一次遍历的所有顶点的堆栈，其第一个元素是根节点。如果在DFT期间遇到一个元素已经在堆栈中，那么就有一个循环。

Tarjan的强连通分量算法的时间复杂度为O(|E| + |V|)。

有关其他算法，请参见维基百科上的强连接组件。

我的方法是做一个拓扑排序，计算访问顶点的数量。如果这个数字小于DAG中的顶点总数，那么就有一个循环。