思路
- 他是先选出各边，然后进行排序，从小到大，下面的代码是存储时就进行升序；
- 使用并查集来进行判断他是否重边，下面代码使用了路径压缩，应该还是比较好懂的；
- 然后就没了，思路就是这么简单；
点击查看代码
import java.io.IOException; import java.util.Scanner; public class Main { static int[] flag; public static void main(String[] args) throws IOException { LinkNode p; Scanner sc = new Scanner(System.in); String str = sc.nextLine(); int n = str.length(); LinkList list = new LinkList(); for(int i=0;i<n;i++){ for (int j = 0; j < n; j++) { int temp = sc.nextInt(); if (i >= j) { if (temp != 0) { p = new LinkNode(i, j, temp); list.Insert_order(p); } } } } list.Print(); kruskal(list,n,str); } //kruskal算法的实现 private static void kruskal(LinkList L,int n,String str) { System.out.println(开始采用kruskal算法选择边：); //继续在此完成代码 int sum=0; LinkNode p=L.head; flag=new int[n]; char[] a=str.toCharArray(); for(int i=0;i<n;i++) { flag[i] = i; } while(p!=null) { if(find(p.start)!=find(p.end)) { System.out.println(a[p.start]+-->+a[p.end]+ = +p.length); sum+=p.length; union(p.start,p.end); } p = p.next; } System.out.println(总权值为：+sum); } private static void union(int x, int y) { x=find(flag[x]); y=find(flag[y]); if (x!=y){ flag[x]=y; } } private static int find(int target) { if (flag[target]==target){ return target; } return find(flag[target]); } public static class LinkNode { public int start; //起点 public int end; //终点 public double length; //长度 public LinkNode next; //下一个结点的引用 //构造方法 public LinkNode(){ this.start = 0; this.end = 0; this.length = 0; this.next = null; } public LinkNode(int start,int end,double length){ this.start = start; this.end = end; this.length = length; } } public static class LinkList { public LinkNode head; //表头结点 //在链表插入是排序 public void Insert_order(LinkNode p){ if(head==null) { head=p; }else { if(p.length<head.length) { p.next=head; head=p; } else { LinkNode current= head; while(current.next!=null &&p.length > current.next.length) { current = current.next; } p.next = current.next; current.next = p; } } } //输出有序链表 public void Print(){ LinkNode p=head.next; System.out.println(有序链表输出如下：); while(p!=null){ System.out.println(p.start+-->+p.end+ = +p.length); p = p.next; } } } }