Kruskal 算法 最小生成树
对于连通的带权图(连通网)G,其生成树也是带权的。生成树T各边的权值总和称为该树的权。
权最小的生成树称为G的最小生成树(Minimum SpannirngTree)。最小生成树可简记为MST。
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/*
Kruskal's algorithm finds a minimum spanning tree for a connected weighted graph.
The program below uses a hard-coded example.
You can change this to match your problem by changing the edges in the graph.
The program uses 3 classes.
The Kruskal class contains the main method.
The Edge class represents an edge.
The KruskalEdges class contains the edges determined by the Kruskal algorithm.
*/
import java.util.TreeSet;
import java.util.Vector;
import java.util.HashSet;
class Edge implements Comparable<Edge>
{
String vertexA, vertexB;
int weight;
public Edge(String vertexA, String vertexB, int weight)
{
this.vertexA = vertexA;
this.vertexB = vertexB;
this.weight = weight;
}
public String getVertexA()
{
return vertexA;
}
public String getVertexB()
{
return vertexB;
}
public int getWeight()
{
return weight;
}
@Override
public String toString()
{
return "(" + vertexA + ", " + vertexB + ") : Weight = " + weight;
}
public int compareTo(Edge edge)
{
//== is not compared so that duplicate values are not eliminated.
return (this.weight < edge.weight) ? -1: 1;
}
}
class KruskalEdges
{
Vector<HashSet<String>> vertexGroups = new Vector<HashSet<String>>();
TreeSet<Edge> kruskalEdges = new TreeSet<Edge>();
public TreeSet<Edge> getEdges()
{
return kruskalEdges;
}
HashSet<String> getVertexGroup(String vertex)
{
for (HashSet<String> vertexGroup : vertexGroups) {
if (vertexGroup.contains(vertex)) {
return vertexGroup;
}
}
return null;
}
/**
* The edge to be inserted has 2 vertices - A and B
* We maintain a vector that contains groups of vertices.
* We first check if either A or B exists in any group
* If neither A nor B exists in any group
* We create a new group containing both the vertices.
* If one of the vertices exists in a group and the other does not
* We add the vertex that does not exist to the group of the other vertex
* If both vertices exist in different groups
* We merge the two groups into one
* All of the above scenarios mean that the edge is a valid Kruskal edge
* In that scenario, we will add the edge to the Kruskal edges
* However, if both vertices exist in the same group
* We do not consider the edge as a valid Kruskal edge
*/
public void insertEdge(Edge edge)
{
String vertexA = edge.getVertexA();
String vertexB = edge.getVertexB();
HashSet<String> vertexGroupA = getVertexGroup(vertexA);
HashSet<String> vertexGroupB = getVertexGroup(vertexB);
if (vertexGroupA == null) {
kruskalEdges.add(edge);
if (vertexGroupB == null) {
HashSet<String> htNewVertexGroup = new HashSet<String>();
htNewVertexGroup.add(vertexA);
htNewVertexGroup.add(vertexB);
vertexGroups.add(htNewVertexGroup);
}
else {
vertexGroupB.add(vertexA);
}
}
else {
if (vertexGroupB == null) {
vertexGroupA.add(vertexB);
kruskalEdges.add(edge);
}
else if (vertexGroupA != vertexGroupB) {
vertexGroupA.addAll(vertexGroupB);
vertexGroups.remove(vertexGroupB);
kruskalEdges.add(edge);
}
}
}
}
public class Kruskal
{
public static void main(String[] args)
{
//TreeSet is used to sort the edges before passing to the algorithm
TreeSet<Edge> edges = new TreeSet<Edge>();
//Sample problem - replace these values with your problem set
edges.add(new Edge("0", "1", 2));
edges.add(new Edge("0", "3", 1));
edges.add(new Edge("1", "2", 3));
edges.add(new Edge("2", "3", 5));
edges.add(new Edge("2", "4", 7));
edges.add(new Edge("3", "4", 6));
edges.add(new Edge("4", "5", 4));
System.out.println("Graph");
KruskalEdges vv = new KruskalEdges();
for (Edge edge : edges) {
System.out.println(edge);
vv.insertEdge(edge);
}
System.out.println("Kruskal algorithm");
int total = 0;
for (Edge edge : vv.getEdges()) {
System.out.println(edge);
total += edge.getWeight();
}
System.out.println("Total weight is " + total);
}
}