Algs4-1.5.12使用路径压缩的quick-union算法

1.5.12使用路径压缩的quick-union算法。根据路径压缩修改quick-union算法(请见1.5.2.3节),在find()方法中添加一个循环来将从p到根节点的路径上的每个触点都连接到根节点。给出一列输入,使该方法能够产生一条长度为4的路径。注意:该算法的所有操作的均摊成本已知为对数级别。
答:
0-1
1-2
2-3
3-4
图片
图片
tinyUF.txt
5
0
1
1
2
2
3
3
4

public class E1d5d12
{
    private int[] id;
    private int count;
    public E1d5d12(int N)
    {
        count=N;
        id=new int[N];
        for (int i=0;i<N;i++)
            id[i]=i;
   }
   
   public int count()
   {return count;}
    
   boolean connected(int p,int q)
   {return find(p)==find(q);}


    public int find(int p)
     {
        int temp=p;
        //find root
        while(p!=id[p]) p=id[p];
        int root=p;
        //set all node's father is root
        p=temp;
         while(root!=id[p])
          {
              temp=id[p];
              id[p]=root;
              p=temp;
          }
          return root;
      }
     
      public void union(int p,int q)
      {
          int pRoot=find(p);
          int qRoot=find(q);
          if(pRoot==qRoot) return;
          id[pRoot]=qRoot;
          count--;
      }

       public static void main(String[] qrgs)
       {
           int N=StdIn.readInt();
           E1d5d12 uf=new E1d5d12(N);
           while (!StdIn.isEmpty())
           {
               int p=StdIn.readInt();
               int q=StdIn.readInt();
               if(uf.connected(p,q)) continue;
               StdOut.printf("p=%d q=%d ",p,q);
               uf.union(p,q);
              
            }//end while
        }//end main
}//end class
原文地址:https://www.cnblogs.com/longjin2018/p/9854765.html