最小树-克鲁斯卡尔

来源:互联网 发布:linux限制ip访问次数 编辑:程序博客网 时间:2024/06/09 18:29
public class Main {static final int MAXVEX =65535;static final int MINVEX =65535;public static void main(String[] args) {MyGrop2 p = new MyGrop2();new Main().MiniSpanTree(p);}static int Find (int parent [] , int f){while(parent[f] >0){f = parent[f]; // 下标 f 指向 parent[f]}return f;} void MiniSpanTree(MyGrop2 G){int i,n,m;EDGE [] edges = new EDGE[MAXVEX];for (i=0 ; i<G.srd.length; i++){edges[i] = new EDGE();edges[i].begin = G.srd[i][0];edges[i].end =G. srd[i][1];edges[i].weight =G.srd[i][2];}//定义变得数组int parent [] =  new int[MAXVEX];//定义parent来判断便于变是否能形成环for( i=0 ;i<G.a ;i++ ){parent[i] = 0;  //初始化}for (i=0 ;i<=G.a ;i++ ){n = Find(parent ,edges[i].begin);m = Find(parent ,edges[i].end);if(n!=m){  //如果n==m 形成环路 parent[n] = m;   //将此节点放入下标为起点parent的数组中,表示此节点存在System.out.println("("+edges[i].begin+")--"+edges[i].weight+"--("+edges[i].end+")");}}}}class EDGE{int begin;int end;int weight;}class MyGrop2{final int MINVEX =65535;int a = 9;int srd [][] = {{4,7,7},{2,8,8},{0,1,10},{0,5,11},{1,8,12},{3,7,16},{1,6,16},{5,6,17},{1,2,18},{6,7,19},{3,4,20},{3,8,21},{2,3,22},{3,6,24},{4,5,26}};}

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