/*** sample Kruskal.java Description:

* kruskal算法的思想是找最小边，且每次找到的边不会和以找出来的边形成环路，利用一个一维数组group存放当前顶点所在连通图标示(每条最小边，属于一个连通图)，直到顶点都找完

* 1.0 YESUN Jul 18,

* 2013 8:48:28 AM Create. ChangeLog:*/

public classKruskal {/*** Description:

*

*@paramargs

* 1.0 YESUN Jul 18, 2013 8:47:10 AM Create. ChangeLog:*/

public static voidmain(String[] args) {int[][] map = new int[][]{

{0,10,MAX,MAX,MAX,11,MAX,MAX,MAX},

{10,0,18,MAX,MAX,MAX,16,MAX,12},

{MAX,MAX,0,22,MAX,MAX,MAX,MAX,8},

{MAX,MAX,22,0,20,MAX,MAX,16,21},

{MAX,MAX,MAX,20,0,26,MAX,7,MAX},

{11,MAX,MAX,MAX,26,0,17,MAX,MAX},

{MAX,16,MAX,MAX,MAX,17,0,19,MAX},

{MAX,MAX,MAX,16,7,MAX,19,0,MAX},

{MAX,12,8,21,MAX,MAX,MAX,MAX,0}

};

kruskal(map);

}static int MAX =Integer.MAX_VALUE;/*** Description: by yesun

*@paramarcs

* 1.0 YESUN Jul 18, 2013 1:42:42 PM Create.

* ChangeLog:*/

public static void kruskal(int[][] arcs) {//顶点个数

int num =arcs.length;//存放对应顶点所在连通图标识

int[] group = new int[num];int sum = 0, n1 = 0, n2 = 0;boolean finished = false;int groupNum = 1;while(!finished) {int min =Integer.MAX_VALUE;//找出所有边中最小值

for(int i = 0; i < num; i++) {for(int j = i+1; j < num; j++) {if(arcs[i][j] > 0 && arcs[i][j]

if (group[i] != group[j] || (group[i] == 0 && group[j] == 0)) {

min=arcs[i][j];

n1=i;

n2=j;

}

}

}

}if(min ==Integer.MAX_VALUE){continue;

}

System.out.println(n1+ " ---> " + n2 + " " +min);

sum+=min;//找到了最小值，设置连通标记

if(group[n1] == 0 && group[n2] == 0){

group[n1]=groupNum;

group[n2]=groupNum;

groupNum++;

}else if(group[n1] > 0 && group[n2] > 0) {int tmp =group[n2];for(int m = 0; m < group.length; m++){if(group[m] ==tmp){

group[m]=group[n1];

}

}

}else{if(group[n1] == 0){

group[n1]=group[n2];

}else{

group[n2]=group[n1];

}

}for(int i = 0; i < group.length; i++) {if(group[i] != group[0]){

finished= false;break;

}else{

finished= true;

}

}if(finished) {break;

}

}

System.out.println(" sum:"+sum);

}

}