Algs4-1.5.13使用路径压缩的加权quick-union算法

1.5.13使用路径压缩的加权quick-union算法。修改加权quick-union算法(算法1.5)，实现如练习1.5.12所述的路径压缩。给出一列输入，使该方法能够产生一棵高度为4的树。注意：访算法的所有操作的均摊成本已知被限制在反Ackermann函数的范围之内，且对于实际应用中可能出现的所有N值均小于5。
答：
1-0
3-2
3-1
5-4
7-6
7-5
7-3
9-8
11-10
11-9
13-12
15-14
15-13
15-11
15-7





tinyUF.txt
16
1
0
3
2
3
1
5
4
7
6
7
5
7
3
9
8
11
10
11
9
13
12
15
14
15
13
15
11
15
7
public class E1d5d13
{
    private int[] id;
    private int[] sz;
    private int count;
    public E1d5d13(int N)
    {
        count=N;
        id=new int[N];
        for (int i=0;i<N;i++)
            id[i]=i;
        //
        sz=new int[N];
        for (int i=0;i<N;i++)
            sz[i]=1;
        //
    }
   
     public int count()
     {return count;}
    
      boolean connected(int p,int q)
      {return find(p)==find(q);}
    
      public int find(int p)
      {
         
          while(p!=id[p]) p=id[p];
          int root=p;
          int temp;
          while(p!=id[p])
          {
              temp=id[p];
              id[p]=root;
              p=temp;
          }
          return root;
      }
      
    
      public void union(int p,int q)
      {
          int i=find(p);
          int j=find(q);
          if(i==j) return;
          if(sz[i]<sz[j])
          {
              id[i]=j;
              sz[j]=sz[j]+sz[i];
           }
          else
          {
              id[j]=i;
              sz[i]=sz[i]+sz[j];
          }
          count--;
          //
          for (int k=0;k<id.length;k++)
              StdOut.printf("%3d",id[k]);
          StdOut.println();
      }
      
       public static void main(String[] qrgs)
       {
           int N=StdIn.readInt();
           E1d5d13 uf=new E1d5d13(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