pyomo：为pyomo中的tsp绘制最佳路线

Question

我仍在学习pyomo，到目前为止，我已经取得了一些进展：这个链接！给出了pyomo中的tsp示例。我复制了以下代码。而且一切正常。但是，我无法打印出最佳路线，有人可以帮忙或者给我一个关于如何打印和绘制最佳路线的思路吗？编码：from pyomo.environ import * from pyomo.opt import SolverFactoryimport pyomo.environn=13distanceMatrix=[[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 = 0model = 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) == 2model.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采用的形式。如果我们看一下这个实现，我们就会看到矩阵x的维数为NxN，域二进制为0或1。考虑元素x [j] [k]等于1的含义。如果j <k或j> k，这意味着什么？

要提取x的值，一种简单的方法可能是

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，您可以根据需要使用x的值。