pyomo:绘制tsp在pyomo中的最佳路径

Question

我还在学习pyomo，到目前为止，我已经取得了一些进展：

这个link！给出了tsp在pyomo中的一个例子。我将代码复制如下。一切都很顺利。然而，我无法打印最佳路线，有人能帮我或给我一个公平的想法如何打印和绘制最佳路线吗？

守则：

from pyomo.environ import * 
from pyomo.opt import SolverFactory
import pyomo.environ

n=13
distanceMatrix=[[0,8,4,10,12,9,15,8,11,5,9,4,10],
[8,0,27,6,8,6,17,10,12,9,8,7,5],
[4,7,0,7,9,5,8,5,4,8,6  ,10,8],
[10,6   ,7,0,6,11,5 ,9,8,12,11,6,9],
[12,8   ,19,6,   0,7,9,6,9,8,4,11,10],
[9,6,5,11,7,0,10,4,3,10,6,5,7],
[15,7   ,8,15,19,10,0,10,9,8,5,9,10],
[8,10   ,5,9,6,4,10,0,11,5,9,6,7],
[11,12,4,8, 19,13,9,11,0, 9,11,11,6],
[5,9,8,12,8,10,8,5,9,0,6,7,5],
   [9,8,6,11,14,6,5,9,11,6,0,10,7],
   [4,7,10,6,31,5,9,6,11,7,10,0,9],
   [10,5,8,9,10,7,10,7,6,5,7,9,0]] 
startCity = 0

model = ConcreteModel()
model.M = Set(initialize=range(1, n+1))
model.N = Set(initialize=range(1, n+1))
model.c = Param(model.N, model.N, initialize=lambda model, i, j:    distanceMatrix[i-1][j-1])
model.x = Var(model.N, model.N, within=Binary)


def obj_rule(model):  

    return sum(model.c[n,j]*model.x[n,j] for n in model.N for j in model.N)
model.obj = Objective(rule=obj_rule,sense=minimize)

def con_rule(model, n):
    return sum(model.x[j,n] for j in model.N if j < n) + sum(model.x[n,j]   for j in model.N if j > n) == 2

model.con = Constraint(model.N, rule=con_rule,doc='constraint1')
opt = SolverFactory("glpk")
results = opt.solve(model)
results.write()
print('Printing Values')
print(value(model.obj))

Answer 1

首先，您必须考虑解决方案的形式。如果我们查看这个实现，我们可以看到矩阵x具有维数NxN和域二进制(0或1)。想想元素xj等于1意味着什么，如果j< k或j>k，这意味着什么？

要提取x的值，一个简单的方法可能是

import numpy as np
N = len(model.N)
x = np.zeros((N,N))
for (i,j), val in model.x.get_values().items():
    x[i-1,j-1] = val

然后，您可以使用x的值，无论您喜欢。