最优二叉查找树

>>算法导论P217

def Optimal_BST(p,q,n):    e=[[0 for col in range(n+1)]for row in range(n+2)] #期望代价    w=[[0 for col in range(n+1)]for row in range(n+2)] #概率总和    root=[[0 for col in range(n+1)]for row in range(n+2)] #根节点    for i in range(1,n+2):        e[i][i-1]=q[i-1]        w[i][i-1]=q[i-1]    for L in range(1,n+1):        for i in range(n-L+2):            j=i+L-1            e[i][j]=float('inf')            w[i][j]=w[i][j-1]+p[j]+q[j]            for r in range(i,j+1): #可用Knuth结果改进算法复杂度                t=e[i][r-1]+e[r+1][j]+w[i][j]                if t<e[i][j]:                    e[i][j]=t                    root[i][j]=r                        Construct_Optimal_BST(root,1,n,n,0)    return 0                   def Construct_Optimal_BST(root,i,j,n,count):    count+=1    if count>(2*n+1):        return     if i==1 and j==n:        print('K'+str(root[i][j])+' is the root')    if i<j:        if root[i][root[i][j]-1]>0 :            print('K'+str(root[i][root[i][j]-1])+' is the left child of K'+str(root[i][j]))        Construct_Optimal_BST(root,i,root[i][j]-1,n,count)        if root[root[i][j]+1][j]>0 :            print('K'+str(root[root[i][j]+1][j])+' is the right child of K'+str(root[i][j]))        Construct_Optimal_BST(root,root[i][j]+1,j,n,count)    if i==j:        print('d'+str(i-1)+' is the left child of K'+str(i))        print('d'+str(i)+' is the right child of K'+str(i))    if i>j:        print('d'+str(j)+' is the right child of K'+str(j))            #testp=[0,0.15,0.1,0.05,0.1,0.2]q=[0.05,0.1,0.05,0.05,0.05,0.1]n=len(p)-1Optimal_BST(p,q,n)