Kruskal 最小生成树算法
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来源：Dennis Gao
链接：http://www.cnblogs.com/gaochundong/p/kruskal_minimum_spanning_tree.html
对于一个给定的连通的无向图 G = (V, E)，希望找到一个无回路的子集 T，T 是 E 的子集，它连接了所有的顶点，且其权值之和为最小。
因为 T 无回路且连接所有的顶点，所以它必然是一棵树，称为生成树（Spanning Tree），因为它生成了图 G。显然，由于树 T 连接了所有的顶点，所以树 T 有 V – 1 条边。一张图 G 可以有很多棵生成树，而把确定权值最小的树 T 的问题称为最小生成树问题（Minimum Spanning Tree）。术语 “最小生成树” 实际上是 “最小权值生成树” 的缩写。
Kruskal 算法提供一种在 O(ElogV) 运行时间确定最小生成树的方案。Kruskal 算法基于贪心算法（Greedy Algorithm）的思想进行设计，其选择的贪心策略就是，每次都选择权重最小的但未形成环路的边加入到生成树中。其算法结构如下：
将所有的边按照权重非递减排序；
选择最小权重的边，判断是否其在当前的生成树中形成了一个环路。如果环路没有形成，则将该边加入树中，否则放弃。
重复步骤 2，直到有 V – 1 条边在生成树中。
上述步骤 2 中使用了 Union-Find 算法来判断是否存在环路。
例如，下面是一个无向连通图 G。
图 G 中包含 9 个顶点和 14 条边，所以期待的最小生成树应包含 (9 – 1) = 8 条边。
首先对所有的边按照权重的非递减顺序排序：
Weight Src Dest
1 7 6
2 8 2
2 6 5
4 0 1
4 2 5
6 8 6
7 2 3
7 7 8
8 0 7
8 1 2
9 3 4
10 5 4
11 1 7
14 3 5
然后从排序后的列表中选择权重最小的边。
1. 选择边 {7, 6}，无环路形成，包含在生成树中。
2. 选择边 {8, 2}，无环路形成，包含在生成树中。
3. 选择边 {6, 5}，无环路形成，包含在生成树中。
4. 选择边 {0, 1}，无环路形成，包含在生成树中。
5. 选择边 {2, 5}，无环路形成，包含在生成树中。
6. 选择边 {8, 6}，有环路形成，放弃。
7. 选择边 {2, 3}，无环路形成，包含在生成树中。
8. 选择边 {7, 8}，有环路形成，放弃。
9. 选择边 {0, 7}，无环路形成，包含在生成树中。
10. 选择边 {1, 2}，有环路形成，放弃。
11. 选择边 {3, 4}，无环路形成，包含在生成树中。
12. 由于当前生成树中已经包含 V – 1 条边，算法结束。
C# 实现的 Kruskal 算法如下。
using System;
using System.Collections.Generic;
using System.Linq;
namespace GraphAlgorithmTesting
{
class Program
{
static void Main(string[] args)
{
Graph g = new Graph(9);
g.AddEdge(0, 1, 4);
g.AddEdge(0, 7, 8);
g.AddEdge(1, 2, 8);
g.AddEdge(1, 7, 11);
g.AddEdge(2, 3, 7);
g.AddEdge(2, 5, 4);
g.AddEdge(8, 2, 2);
g.AddEdge(3, 4, 9);
g.AddEdge(3, 5, 14);
g.AddEdge(5, 4, 10);
g.AddEdge(6, 5, 2);
g.AddEdge(8, 6, 6);
g.AddEdge(7, 6, 1);
g.AddEdge(7, 8, 7);
Console.WriteLine();
Console.WriteLine("Graph Vertex Count : {0}", g.VertexCount);
Console.WriteLine("Graph Edge Count : {0}", g.EdgeCount);
Console.WriteLine();
Console.WriteLine("Is there cycle in graph: {0}", g.HasCycle());
Console.WriteLine();
Edge[] mst = g.Kruskal();
Console.WriteLine("MST Edges:");
foreach (var edge in mst)
{
Console.WriteLine("\t{0}", edge);
}
Console.ReadKey();
}
class Edge
{
public Edge(int begin, int end, int weight)
{
this.Begin = begin;
this.End = end;
this.Weight = weight;
}
public int Begin { get; private set; }
public int End { get; private set; }
public int Weight { get; private set; }
public override string ToString()
{
return string.Format(
"Begin[{0}], End[{1}], Weight[{2}]",
Begin, End, Weight);
}
}
class Subset
{
public int Parent { get; set; }
public int Rank { get; set; }
}
class Graph
{
private Dictionary<int, List<Edge>> _adjacentEdges
= new Dictionary<int, List<Edge>>();
public Graph(int vertexCount)
{
this.VertexCount = vertexCount;
}
public int VertexCount { get; private set; }
public IEnumerable<int> Vertices { get { return _adjacentEdges.Keys; } }
public IEnumerable<Edge> Edges
{
get { return _adjacentEdges.Values.SelectMany(e => e); }
}
public int EdgeCount { get { return this.Edges.Count(); } }
public void AddEdge(int begin, int end, int weight)
{
if (!_adjacentEdges.ContainsKey(begin))
{
var edges = new List<Edge>();
_adjacentEdges.Add(begin, edges);
}
_adjacentEdges[begin].Add(new Edge(begin, end, weight));
}
private int Find(Subset[] subsets, int i)
{
// find root and make root as parent of i (path compression)
if (subsets[i].Parent != i)
subsets[i].Parent = Find(subsets, subsets[i].Parent);
return subsets[i].Parent;
}
private void Union(Subset[] subsets, int x, int y)
{
int xroot = Find(subsets, x);
int yroot = Find(subsets, y);
// Attach smaller rank tree under root of high rank tree
// (Union by Rank)
if (subsets[xroot].Rank < subsets[yroot].Rank)
subsets[xroot].Parent = yroot;
else if (subsets[xroot].Rank > subsets[yroot].Rank)
subsets[yroot].Parent = xroot;
// If ranks are same, then make one as root and increment
// its rank by one
else
{
subsets[yroot].Parent = xroot;
subsets[xroot].Rank++;
}
}
public bool HasCycle()
{
Subset[] subsets = new Subset[VertexCount];
for (int i = 0; i < subsets.Length; i++)
{
subsets[i] = new Subset();
subsets[i].Parent = i;
subsets[i].Rank = 0;
}
// Iterate through all edges of graph, find subset of both
// vertices of every edge, if both subsets are same,
// then there is cycle in graph.
foreach (var edge in this.Edges)
{
int x = Find(subsets, edge.Begin);
int y = Find(subsets, edge.End);
if (x == y)
{
return true;
}
Union(subsets, x, y);
}
return false;
}
public Edge[] Kruskal()
{
// This will store the resultant MST
Edge[] mst = new Edge[VertexCount - 1];
// Step 1: Sort all the edges in non-decreasing order of their weight
// If we are not allowed to change the given graph, we can create a copy of
// array of edges
var sortedEdges = this.Edges.OrderBy(t => t.Weight);
var enumerator = sortedEdges.GetEnumerator();
// Allocate memory for creating V ssubsets
// Create V subsets with single elements
Subset[] subsets = new Subset[VertexCount];
for (int i = 0; i < subsets.Length; i++)
{
subsets[i] = new Subset();
subsets[i].Parent = i;
subsets[i].Rank = 0;
}
// Number of edges to be taken is equal to V-1
int e = 0;
while (e < VertexCount - 1)
{
// Step 2: Pick the smallest edge. And increment the index
// for next iteration
Edge nextEdge;
if (enumerator.MoveNext())
{
nextEdge = enumerator.Current;
int x = Find(subsets, nextEdge.Begin);
int y = Find(subsets, nextEdge.End);
// If including this edge does't cause cycle, include it
// in result and increment the index of result for next edge
if (x != y)
{
mst[e++] = nextEdge;
Union(subsets, x, y);
}
else
{
// Else discard the nextEdge
}
}
}
return mst;
}
}
}
}
输出结果如下：
参考资料
Connectivity in a directed graph
Strongly Connected Components
Tarjan’s Algorithm to find Strongly Connected Components