KRUSKAL算法

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图的KRUSKAL算法是用以求解最小生成树的一种有效算法，每次从待选的边中选出最短的那一条边，加入到已有的树中，读者可以与PRIM算法相比较他们之间的异同，下面是代码，该代码利用堆排序以求能在较短时间内取出最短的那一条边。

package oj;class Edge{int x1;int x2;int len;public Edge(int x1, int x2, int len) {this.x1 = x1;this.x2 = x2;this.len = len;}public int getX1() {return x1;}public void setX1(int x1) {this.x1 = x1;}public int getX2() {return x2;}public void setX2(int x2) {this.x2 = x2;}public int getLen() {return len;}public void setLen(int len) {this.len = len;}}public class KRUSKAL {public static void kruskal(Edge2[] edges,int[] point){preSortByLen(edges);int[] p=makeSet(point);int[] rank=initRank(point);Edge2[] T=new Edge2[point.length-1]; //最小生成树的边数是定点数-1int k=0;                           //该指针用来表示edges中的下一条边int t=0;while(t<point.length-1){Edge2 e=edges[k++];if(find(e.getX1(),p)!=find(e.getX2(),p)){T[t++]=e;union(e.getX1(),e.getX2(),p,rank);}}int sum=0;for(int j=0;j<T.length;j++){System.out.println(j+1+" :"+T[j].getLen());}}//初始化建堆，对每个顶点来说他的父节点是他本身public static int[] makeSet(int[] point){int[] p=new int[point.length];for(int i=0;i<point.length;i++){p[point[i]]=point[i];}return p;}public static int find(int x,int[] p){int y=x;while(p[y]!=y){y=p[y];}int root=y;return root;}public static int[] initRank(int[] point){int[] rank=new int[point.length];for(int i=0;i<point.length;i++){rank[point[i]]=1;}return rank;}public static void union(int x,int y,int p[],int rank[]){int r1=find(x,p);int r2=find(y,p);if(rank[r1]<=rank[r2]){p[r1]=r2;if(rank[r1]==rank[r2]){rank[r2]++;}}else{p[r2]=r1;}}public static void preSortByLen(Edge2[] edges){int len=edges.length;for(int i=1;i<=len-1;i++){for(int j=edges.length-1;j>=i;j--){if(edges[j].getLen()<edges[j-1].getLen()){Edge2 temp=edges[j-1];edges[j-1]=edges[j];edges[j]=temp;}}}}public static void main(String args[]){Edge2 e1=new Edge2(0,1,1);Edge2 e2=new Edge2(0,2,2);Edge2 e3=new Edge2(1,2,6);Edge2 e4=new Edge2(1,3,11);Edge2 e5=new Edge2(2,3,9);Edge2 e6=new Edge2(2,4,13);Edge2 e7=new Edge2(3,4,7);Edge2 e8=new Edge2(3,5,3);Edge2 e9=new Edge2(4,5,4);Edge2 edges[]={e1,e2,e3,e4,e5,e6,e7,e8,e9};int[] point=new int[6];for(int i=0;i<point.length;i++){point[i]=i;}kruskal(edges,point);}}

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