Kruskal 最小生成树算法
原文 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: