假设我有一个图表，其中节点存储在排序列表中。我现在想要对这个图进行拓扑排序，同时保持拓扑顺序未定义的原始顺序。 对此有什么好的算法吗？
答案 0 :(得分：16)
一种可能性是计算按字典顺序排列的最小拓扑顺序。该算法用于维护一个优先级队列，该队列包含有效度（在尚未处理的节点上）为零的节点。使用最小标签反复出列节点，将其附加到顺序，减少其后继者的有效入度，将现在具有度数为零的那些排队。这在btilly的例子中产生1234567890，但通常不会最小化反转。
我喜欢这个算法的属性是输出有一个清晰的定义，显然只满足一个顺序，并且只要有反转（节点x出现在节点y之后，即使x
答案 1 :(得分：4)
问题是双重的：
在经过多次错误和试验之后，我想出了一个类似于冒泡排序但具有拓扑顺序标准的简单算法。
我在完整图表上使用完整的边缘组合彻底测试了算法，因此可以认为它已被证实。
可以容忍循环依赖性并根据元素的原始顺序进行解析。结果顺序是完美的，代表最接近的匹配。
以下是C＃中的源代码：
static class TopologicalSort
{
/// <summary>
/// Delegate definition for dependency function.
/// </summary>
/// <typeparam name="T">The type.</typeparam>
/// <param name="a">The A.</param>
/// <param name="b">The B.</param>
/// <returns>
/// Returns <c>true</c> when A depends on B. Otherwise, <c>false</c>.
/// </returns>
public delegate bool TopologicalDependencyFunction<in T>(T a, T b);
/// <summary>
/// Sorts the elements of a sequence in dependency order according to comparison function with Gapotchenko algorithm.
/// The sort is stable. Cyclic dependencies are tolerated and resolved according to original order of elements in sequence.
/// </summary>
/// <typeparam name="T">The type of the elements of source.</typeparam>
/// <param name="source">A sequence of values to order.</param>
/// <param name="dependencyFunction">The dependency function.</param>
/// <param name="equalityComparer">The equality comparer.</param>
/// <returns>The ordered sequence.</returns>
public static IEnumerable<T> StableOrder<T>(
IEnumerable<T> source,
TopologicalDependencyFunction<T> dependencyFunction,
IEqualityComparer<T> equalityComparer)
{
if (source == null)
throw new ArgumentNullException("source");
if (dependencyFunction == null)
throw new ArgumentNullException("dependencyFunction");
if (equalityComparer == null)
throw new ArgumentNullException("equalityComparer");
var graph = DependencyGraph<T>.TryCreate(source, dependencyFunction, equalityComparer);
if (graph == null)
return source;
var list = source.ToList();
int n = list.Count;
Restart:
for (int i = 0; i < n; ++i)
{
for (int j = 0; j < i; ++j)
{
if (graph.DoesXHaveDirectDependencyOnY(list[j], list[i]))
{
bool jOnI = graph.DoesXHaveTransientDependencyOnY(list[j], list[i]);
bool iOnJ = graph.DoesXHaveTransientDependencyOnY(list[i], list[j]);
bool circularDependency = jOnI && iOnJ;
if (!circularDependency)
{
var t = list[i];
list.RemoveAt(i);
list.Insert(j, t);
goto Restart;
}
}
}
}
return list;
}
/// <summary>
/// Sorts the elements of a sequence in dependency order according to comparison function with Gapotchenko algorithm.
/// The sort is stable. Cyclic dependencies are tolerated and resolved according to original order of elements in sequence.
/// </summary>
/// <typeparam name="T">The type of the elements of source.</typeparam>
/// <param name="source">A sequence of values to order.</param>
/// <param name="dependencyFunction">The dependency function.</param>
/// <returns>The ordered sequence.</returns>
public static IEnumerable<T> StableOrder<T>(
IEnumerable<T> source,
TopologicalDependencyFunction<T> dependencyFunction)
{
return StableOrder(source, dependencyFunction, EqualityComparer<T>.Default);
}
sealed class DependencyGraph<T>
{
private DependencyGraph()
{
}
public IEqualityComparer<T> EqualityComparer
{
get;
private set;
}
public sealed class Node
{
public int Position
{
get;
set;
}
List<T> _Children = new List<T>();
public IList<T> Children
{
get
{
return _Children;
}
}
}
public IDictionary<T, Node> Nodes
{
get;
private set;
}
public static DependencyGraph<T> TryCreate(
IEnumerable<T> source,
TopologicalDependencyFunction<T> dependencyFunction,
IEqualityComparer<T> equalityComparer)
{
var list = source as IList<T>;
if (list == null)
list = source.ToArray();
int n = list.Count;
if (n < 2)
return null;
var graph = new DependencyGraph<T>();
graph.EqualityComparer = equalityComparer;
graph.Nodes = new Dictionary<T, Node>(n, equalityComparer);
bool hasDependencies = false;
for (int position = 0; position < n; ++position)
{
var element = list[position];
Node node;
if (!graph.Nodes.TryGetValue(element, out node))
{
node = new Node();
node.Position = position;
graph.Nodes.Add(element, node);
}
foreach (var anotherElement in list)
{
if (equalityComparer.Equals(element, anotherElement))
continue;
if (dependencyFunction(element, anotherElement))
{
node.Children.Add(anotherElement);
hasDependencies = true;
}
}
}
if (!hasDependencies)
return null;
return graph;
}
public bool DoesXHaveDirectDependencyOnY(T x, T y)
{
Node node;
if (Nodes.TryGetValue(x, out node))
{
if (node.Children.Contains(y, EqualityComparer))
return true;
}
return false;
}
sealed class DependencyTraverser
{
public DependencyTraverser(DependencyGraph<T> graph)
{
_Graph = graph;
_VisitedNodes = new HashSet<T>(graph.EqualityComparer);
}
DependencyGraph<T> _Graph;
HashSet<T> _VisitedNodes;
public bool DoesXHaveTransientDependencyOnY(T x, T y)
{
if (!_VisitedNodes.Add(x))
return false;
Node node;
if (_Graph.Nodes.TryGetValue(x, out node))
{
if (node.Children.Contains(y, _Graph.EqualityComparer))
return true;
foreach (var i in node.Children)
{
if (DoesXHaveTransientDependencyOnY(i, y))
return true;
}
}
return false;
}
}
public bool DoesXHaveTransientDependencyOnY(T x, T y)
{
var traverser = new DependencyTraverser(this);
return traverser.DoesXHaveTransientDependencyOnY(x, y);
}
}
}
一个小样本申请：
class Program
{
static bool DependencyFunction(char a, char b)
{
switch (a + " depends on " + b)
{
case "A depends on B":
return true;
case "B depends on D":
return true;
default:
return false;
}
}
static void Main(string[] args)
{
var source = "ABCDEF";
var result = TopologicalSort.StableOrder(source.ToCharArray(), DependencyFunction);
Console.WriteLine(string.Concat(result));
}
}
给定输入元素{A，B，C，D，E，F}，其中A取决于B而B取决于D，输出为{D，B，A，C，E，F}。
<强>更新 我写了关于稳定拓扑排序目标，算法及其校对的a small article。希望这能给出更多解释，对开发人员和研究人员有用。
答案 2 :(得分：3)
您没有足够的条件来指定您要查找的内容。例如，考虑一个带有两个有向分量的图。
1 -> 2 -> 3 -> 4 -> 5
6 -> 7 -> 8 -> 9 -> 0
您更喜欢以下哪种排序？
6, 7, 8, 9, 0, 1, 2, 3, 4, 5
1, 2, 3, 4, 5, 6, 7, 8, 9, 0
第一个结果是通过将最低节点尽可能靠近列表头部来打破所有联系。因此0胜。第二个结果是试图最小化A＆lt; B和B出现在拓扑排序中的A之前。两者都是合理的答案。第二个可能更令人愉快。
我可以轻松地为第一个算法生成算法。首先，获取最低节点，然后进行广度优先搜索以找到到最短根节点的距离。如果存在平局，请确定可能出现在这样一条最短路径上的节点集。获取该集合中的最低节点，并将最佳路径从它放置到根，然后将最佳路径从我们开始的最低节点放置到它。搜索尚未在拓扑排序中的下一个最低节点，然后继续。
为更令人愉悦的版本生成算法似乎要困难得多。请参阅http://en.wikipedia.org/wiki/Feedback_arc_set了解相关问题，该问题强烈暗示它实际上是NP完全的。
答案 3 :(得分：2)
这是拓扑排序的一种简单迭代方法：不断删除具有度0的节点及其边缘。
要获得稳定版本，只需修改为：连续删除具有in-degree 0的最小索引节点及其边缘。
在伪python中：
# N is the number of nodes, labeled 0..N-1
# edges[i] is a list of nodes j, corresponding to edges (i, j)
inDegree = [0] * N
for i in range(N):
for j in edges[i]:
inDegree[j] += 1
# Now we maintain a "frontier" of in-degree 0 nodes.
# We take the smallest one until the frontier is exhausted.
# Note: You could use a priority queue / heap instead of a list,
# giving O(NlogN) runtime. This naive implementation is
# O(N^2) worst-case (when the order is very ambiguous).
frontier = []
for i in range(N):
if inDegree[i] == 0:
frontier.append(i)
order = []
while frontier:
i = min(frontier)
frontier.remove(i)
for j in edges[i]:
inDegree[j] -= 1
if inDegree[j] == 0:
frontier.append(j)
# Done - order is now a list of the nodes in topological order,
# with ties broken by original order in the list.
答案 4 :(得分：0)
将“稳定拓扑排序”解释为DAG的线性化，使得拓扑顺序无关紧要的线性化范围按字典顺序排序。这可以通过DFS线性化方法来解决，修改是按字典顺序访问节点。
我有一个Python Digraph类，其线性化方法如下所示：
def linearize_as_needed(self):
if self.islinearized:
return
# Algorithm: DFS Topological sort
# https://en.wikipedia.org/wiki/Topological_sorting#Depth-first_search
temporary = set()
permanent = set()
L = [ ]
def visit(vertices):
for vertex in sorted(vertices, reverse=True):
if vertex in permanent:
pass
elif vertex in temporary:
raise NotADAG
else:
temporary.add(vertex)
if vertex in self.arrows:
visit(self.arrows[vertex])
L.append(vertex)
temporary.remove(vertex)
permanent.add(vertex)
# print('visit: {} => {}'.format(vertices, L))
visit(self.vertices)
self._linear = list(reversed(L))
self._iter = iter(self._linear)
self.islinearized = True
下面
self.vertices
是所有顶点的集合，
self.arrows
将邻接关系保存为左节点的dict到右节点集。
答案 5 :(得分：0)
depth-first search algorithm on Wikipedia为我工作：
const assert = chai.assert;
const stableTopologicalSort = ({
edges,
nodes
}) => {
// https://en.wikipedia.org/wiki/Topological_sorting#Depth-first_search
const result = [];
const marks = new Map();
const visit = node => {
if (marks.get(node) !== `permanent`) {
assert.notEqual(marks.get(node), `temporary`, `not a DAG`);
marks.set(node, `temporary`);
edges.filter(([, to]) => to === node).forEach(([from]) => visit(from));
marks.set(node, `permanent`);
result.push(node);
}
};
nodes.forEach(visit);
return result;
};
const graph = {
edges: [
[5, 11],
[7, 11],
[3, 8],
[11, 2],
[11, 9],
[11, 10],
[8, 9],
[3, 10]
],
nodes: [2, 3, 5, 7, 8, 9, 10, 11]
};
assert.deepEqual(stableTopologicalSort(graph), [5, 7, 11, 2, 3, 8, 9, 10]);<script src="https://cdnjs.cloudflare.com/ajax/libs/chai/4.2.0/chai.min.js"></script>