当一个部署项目包含第二个部署项目的项目输出，而第二个项目包含第一个项目的输出时，通常会发生此错误。

我有一个检查循环依赖的方法。在输入中，我们有一个包含例如 <"A", < "B", "C" >> 的字典。和 <"B", < "A", "D" >> ，这意味着 A取决于 B和 C我们与 A->B 有循环依赖关系.

但通常情况更复杂，有一个依赖链。 如何修改此方法以找到依赖链？例如，我想要一个包含链 A->B->A 的变量, 而不是类 A与类 B 有冲突.

private void FindDependency(IDictionary<string, IEnumerable<string>> serviceDependence)

最佳答案

在图中查找循环的一种简单方法是使用递归深度优先图着色算法，其中节点被标记为“正在访问”或“已访问”。如果，当访问一个节点时，发现它已经处于“正在访问”状态，那么你就有了一个循环。可以跳过标记为“已访问”的节点。例如:

public class DependencyExtensions
{
    enum VisitState
    {
        NotVisited,
        Visiting,
        Visited
    };

    public static TValue ValueOrDefault<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TKey key, TValue defaultValue)
    {
        TValue value;
        if (dictionary.TryGetValue(key, out value))
            return value;
        return defaultValue;
    }

    static void DepthFirstSearch<T>(T node, Func<T, IEnumerable<T>> lookup, List<T> parents, Dictionary<T, VisitState> visited, List<List<T>> cycles)
    {
        var state = visited.ValueOrDefault(node, VisitState.NotVisited);
        if (state == VisitState.Visited)
            return;
        else if (state == VisitState.Visiting)
        {
            // Do not report nodes not included in the cycle.
            cycles.Add(parents.Concat(new[] { node }).SkipWhile(parent => !EqualityComparer<T>.Default.Equals(parent, node)).ToList());
        }
        else
        {
            visited[node] = VisitState.Visiting;
            parents.Add(node);
            foreach (var child in lookup(node))
                DepthFirstSearch(child, lookup, parents, visited, cycles);
            parents.RemoveAt(parents.Count - 1);
            visited[node] = VisitState.Visited;
        }
    }

    public static List<List<T>> FindCycles<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> edges)
    {
        var cycles = new List<List<T>>();
        var visited = new Dictionary<T, VisitState>();
        foreach (var node in nodes)
            DepthFirstSearch(node, edges, new List<T>(), visited, cycles);
        return cycles;
    }

    public static List<List<T>> FindCycles<T, TValueList>(this IDictionary<T, TValueList> listDictionary)
        where TValueList : class, IEnumerable<T>
    {
        return listDictionary.Keys.FindCycles(key => listDictionary.ValueOrDefault(key, null) ?? Enumerable.Empty<T>());
    }
}

然后，您可以像这样使用它:

        var serviceDependence = new Dictionary<string, List<string>>
        {
            { "A", new List<string> { "A" }},
            { "B", new List<string> { "C", "D" }},
            { "D", new List<string> { "E" }},
            { "E", new List<string> { "F", "Q" }},
            { "F", new List<string> { "D" }},
        };
        var cycles = serviceDependence.FindCycles();
        Debug.WriteLine(JsonConvert.SerializeObject(cycles, Formatting.Indented));
        foreach (var cycle in cycles)
        {
            serviceDependence[cycle[cycle.Count - 2]].Remove(cycle[cycle.Count - 1]);
        }
        Debug.Assert(serviceDependence.FindCycles().Count == 0);

更新

您的问题已更新为请求用于查找循环依赖项的“最有效算法”。原始答案中的代码是递归的，因此对于数千级深度的依赖链，有可能出现 StackOverflowException。这是一个带有显式堆栈变量的非递归版本:

public static class DependencyExtensions
{
    enum VisitState
    {
        NotVisited,
        Visiting,
        Visited
    };

    public static TValue ValueOrDefault<TKey, TValue>(this IDictionary<TKey, TValue> dictionary, TKey key, TValue defaultValue)
    {
        TValue value;
        if (dictionary.TryGetValue(key, out value))
            return value;
        return defaultValue;
    }

    private static void TryPush<T>(T node, Func<T, IEnumerable<T>> lookup, Stack<KeyValuePair<T, IEnumerator<T>>> stack, Dictionary<T, VisitState> visited, List<List<T>> cycles)
    {
        var state = visited.ValueOrDefault(node, VisitState.NotVisited);
        if (state == VisitState.Visited)
            return;
        else if (state == VisitState.Visiting)
        {
            Debug.Assert(stack.Count > 0);
            var list = stack.Select(pair => pair.Key).TakeWhile(parent => !EqualityComparer<T>.Default.Equals(parent, node)).ToList();
            list.Add(node);
            list.Reverse();
            list.Add(node);
            cycles.Add(list);
        }
        else
        {
            visited[node] = VisitState.Visiting;
            stack.Push(new KeyValuePair<T, IEnumerator<T>>(node, lookup(node).GetEnumerator()));
        }
    }

    static List<List<T>> FindCycles<T>(T root, Func<T, IEnumerable<T>> lookup, Dictionary<T, VisitState> visited)
    {
        var stack = new Stack<KeyValuePair<T, IEnumerator<T>>>();
        var cycles = new List<List<T>>();

        TryPush(root, lookup, stack, visited, cycles);
        while (stack.Count > 0)
        {
            var pair = stack.Peek();
            if (!pair.Value.MoveNext())
            {
                stack.Pop();                    
                visited[pair.Key] = VisitState.Visited;
                pair.Value.Dispose();
            }
            else
            {
                TryPush(pair.Value.Current, lookup, stack, visited, cycles);
            }
        }
        return cycles;
    }

    public static List<List<T>> FindCycles<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> edges)
    {
        var cycles = new List<List<T>>();
        var visited = new Dictionary<T, VisitState>();
        foreach (var node in nodes)
            cycles.AddRange(FindCycles(node, edges, visited));
        return cycles;
    }

    public static List<List<T>> FindCycles<T, TValueList>(this IDictionary<T, TValueList> listDictionary)
        where TValueList : class, IEnumerable<T>
    {
        return listDictionary.Keys.FindCycles(key => listDictionary.ValueOrDefault(key, null) ?? Enumerable.Empty<T>());
    }
}

这在 N*log(N) + E 时应该相当有效，其中 N 是节点数，E 是节点数的边缘。 Log(N) 来自构建visited 哈希表，可以通过让每个节点记住它的VisitState 来消除。这似乎是合理的表现；在下面的测试工具中，在 10000 个节点中找到平均长度为 4393 的 17897 个循环，总依赖项为 125603，所需时间约为 10.2 秒:

public class TestClass
{
    public static void TestBig()
    {
        var elapsed = TestBig(10000);
        Debug.WriteLine(elapsed.ToString());
    }

    static string GetName(int i)
    {
        return "ServiceDependence" + i.ToString();
    }

    public static TimeSpan TestBig(int count)
    {
        var serviceDependence = new Dictionary<string, List<string>>();
        for (int iItem = 0; iItem < count; iItem++)
        {
            var name = GetName(iItem);
            // Add several forward references.
            for (int iRef = iItem - 1; iRef > 0; iRef = iRef / 2)
                serviceDependence.Add(name, GetName(iRef));
            // Add some backwards references.
            if (iItem > 0 && (iItem % 5 == 0))
                serviceDependence.Add(name, GetName(iItem + 5));
        }

        // Add one backwards reference that will create some extremely long cycles.
        serviceDependence.Add(GetName(1), GetName(count - 1));

        List<List<string>> cycles;

        var stopwatch = new Stopwatch();
        stopwatch.Start();
        try
        {
            cycles = serviceDependence.FindCycles();
        }
        finally
        {
            stopwatch.Stop();
        }

        var elapsed = stopwatch.Elapsed;

        var averageLength = cycles.Average(l => (double)l.Count);
        var total = serviceDependence.Values.Sum(l => l.Count);

        foreach (var cycle in cycles)
        {
            serviceDependence[cycle[cycle.Count - 2]].Remove(cycle[cycle.Count - 1]);
        }
        Debug.Assert(serviceDependence.FindCycles().Count == 0);

        Console.WriteLine(string.Format("Time to find {0} cycles of average length {1} in {2} nodes with {3} total dependencies: {4}", cycles.Count, averageLength, count, total, elapsed));
        Console.ReadLine();
        System.Environment.Exit(0);

        return elapsed;
    }
}

关于c# - 在C#中，如何找到循环依赖链？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/29703972/