首先，你并不是真的想要找出所有的循环因为如果有1个，那么就会有无穷多个循环。比如A-B-A, A-B-A- b - a等等。或者可以将2个循环组合成一个8-like循环等等……有意义的方法是寻找所有所谓的简单循环——那些除了开始/结束点之外不交叉的循环。如果你愿意，你可以生成简单循环的组合。

One of the baseline algorithms for finding all simple cycles in a directed graph is this: Do a depth-first traversal of all simple paths (those that do not cross themselves) in the graph. Every time when the current node has a successor on the stack a simple cycle is discovered. It consists of the elements on the stack starting with the identified successor and ending with the top of the stack. Depth first traversal of all simple paths is similar to depth first search but you do not mark/record visited nodes other than those currently on the stack as stop points.

The brute force algorithm above is terribly inefficient and in addition to that generates multiple copies of the cycles. It is however the starting point of multiple practical algorithms which apply various enhancements in order to improve performance and avoid cycle duplication. I was surprised to find out some time ago that these algorithms are not readily available in textbooks and on the web. So I did some research and implemented 4 such algorithms and 1 algorithm for cycles in undirected graphs in an open source Java library here : http://code.google.com/p/niographs/ .

顺便说一句，因为我提到了无向图:它们的算法是不同的。构建一棵生成树，然后每一条不属于树的边与树中的一些边一起形成一个简单的循环。这样发现的循环形成了所谓的循环基。所有的简单循环都可以通过组合两个或多个不同的基循环来找到。更多细节请参见:http://dspace.mit.edu/bitstream/handle/1721.1/68106/FTL_R_1982_07.pdf。

DFS c++版本的伪代码在二楼的答案:

void findCircleUnit(int start, int v, bool* visited, vector<int>& path) {
    if(visited[v]) {
        if(v == start) {
            for(auto c : path)
                cout << c << " ";
            cout << endl;
            return;
        }
        else 
            return;
    }
    visited[v] = true;
    path.push_back(v);
    for(auto i : G[v])
        findCircleUnit(start, i, visited, path);
    visited[v] = false;
    path.pop_back();
}

http://www.me.utexas.edu/~bard/IP/Handouts/cycles.pdf

关于你关于排列周期的问题，请在这里阅读更多: https://www.codechef.com/problems/PCYCLE

您可以尝试以下代码(输入大小和数字number):

# include<cstdio>
using namespace std;

int main()
{
    int n;
    scanf("%d",&n);

    int num[1000];
    int visited[1000]={0};
    int vindex[2000];
    for(int i=1;i<=n;i++)
        scanf("%d",&num[i]);

    int t_visited=0;
    int cycles=0;
    int start=0, index;

    while(t_visited < n)
    {
        for(int i=1;i<=n;i++)
        {
            if(visited[i]==0)
            {
                vindex[start]=i;
                visited[i]=1;
                t_visited++;
                index=start;
                break;
            }
        }
        while(true)
        {
            index++;
            vindex[index]=num[vindex[index-1]];

            if(vindex[index]==vindex[start])
                break;
            visited[vindex[index]]=1;
            t_visited++;
        }
        vindex[++index]=0;
        start=index+1;
        cycles++;
    }

    printf("%d\n",cycles,vindex[0]);

    for(int i=0;i<(n+2*cycles);i++)
    {
        if(vindex[i]==0)
            printf("\n");
        else
            printf("%d ",vindex[i]);
    }
}

深度优先搜索和回溯应该在这里工作。 保存一个布尔值数组，以跟踪您以前是否访问过某个节点。如果您没有新节点可访问(不涉及已经访问过的节点)，那么只需返回并尝试不同的分支。

如果你有一个邻接表来表示图，DFS很容易实现。例如adj[A] = {B,C}表示B和C是A的子结点。

例如，下面的伪代码。“start”是开始的节点。

dfs(adj,node,visited):  
  if (visited[node]):  
    if (node == start):  
      "found a path"  
    return;  
  visited[node]=YES;  
  for child in adj[node]:  
    dfs(adj,child,visited)
  visited[node]=NO;

用开始节点调用上面的函数:

visited = {}
dfs(adj,start,visited)

澄清:

Strongly Connected Components will find all subgraphs that have at least one cycle in them, not all possible cycles in the graph. e.g. if you take all strongly connected components and collapse/group/merge each one of them into one node (i.e. a node per component), you'll get a tree with no cycles (a DAG actually). Each component (which is basically a subgraph with at least one cycle in it) can contain many more possible cycles internally, so SCC will NOT find all possible cycles, it will find all possible groups that have at least one cycle, and if you group them, then the graph will not have cycles. to find all simple cycles in a graph, as others mentioned, Johnson's algorithm is a candidate.