我已经用python实现了docplex模型，一切都很顺利：问题公式，变量，约束条件和目标函数。当我试图计算两个圆心之间的欧几里得距离时提出的问题，简单地numpy可以做np.sqrt（（x2-x1）**2 +（y2-y1）**2）。请告诉我如何解决这个问题，我无法找到欧几里得距离的二次方程，而不使用平方根。
附上工作代码，您可以在其中看到约束1的问题。

from docplex.mp.model import Model
import numpy as np

def packing_cplex(CDA, R, H, items):

  # continuous variables x,y range from -R to +R
  x = [ CDA.continuous_var(name="x{}:".format(i), lb=-R, ub=R)
        for i in range(len(items))]

  y = [ CDA.continuous_var(name="y{}:".format(i), lb=-R, ub=R)
        for i in range(len(items))]
  
  # integer variable z range from 0 to H 
  z = [ CDA.integer_var(name="z{}:".format(i), lb=0, ub=H)
        for i in range(len(items))]
  
  # indicator denotes whether an item is packed into the container
  # i=1, .., n, values 1/0
  d = [ CDA.binary_var(name="d{}:".format(i))
         for i in range(len(items))]

  # sqrt root issue--> np.sqrt((x[i]**2 + y[i]**2))
  # 1.constraint packed items and container radius
  CDA.add_quadratic_constraints(R >= (items[i][0] + (x[i]**2 + y[i]**2)**1 )
                                      for i in range(len(items)))
    
  # 2.constraint packed items and container height
  for i in range(len(items)):                  
    CDA.add( CDA.if_then( d[i]==1, H >= ( items[i][1] + z[i] ) ) )
    
  # objective maximise max∑_(i=0 .. n-1 (π * ri^2 * hi * di)
  CDA.set_objective("max", np.sum([d[i] * np.pi * items[i][0]**2 * items[i][1]
                                                for i in range(len(items))]))

# radius and height of cylinder container  
R, H = 3, 2
volume = np.pi * R**2 * H

# pack, [(ri,hi), ... ([rn,hn]) where ri/hi is radius/height of item
items = [(1,2),(1,2),(1,2),(1,2),(1,2),(1,2),(1,2)]
CDA = Model(name='CDA')
packing_cplex(CDA, R, H, items)
CDA.print_information()
solution = CDA.solve()
utilization = solution._objective / volume
print('Utilization (%) ', utilization)

我期待平方根的替代解