python实现cplex求解tsp问题

TSP问题是旅行商问题，即在给定的一组城市中，旅行商要找到一条最短的路径，使得每个城市恰好被访问一次，最后回到起点。下面是利用IBM的Cplex求解TSP问题的Python代码：

import sys
import cplex
from cplex.exceptions import CplexError
import numpy as np

def solve_tsp(cities):
    n = len(cities)
    dist = np.zeros((n, n))
    for i in range(n):
        for j in range(n):
            dist[i][j] = np.linalg.norm(np.array(cities[i])-np.array(cities[j]))

    my_obj = dist.flatten()
    my_colnames = ["x"+str(i) for i in range(n*n)]
    my_lb = [0.0]*n*n
    my_ub = [1.0]*n*n
    my_ctype = "I"*n*n

    my_rhs = [1.0]*n + [n-1.0]*n
    my_sense = "E"*n + "L"*n

    rows = []
    for i in range(n):
        row = []
        for j in range(n):
            row.append(i*n+j)
        rows.append(cplex.SparsePair(ind=row, val=[1.0]*n))

    for i in range(n):
        row = []
        for j in range(n):
            row.append(j*n+i)
        rows.append(cplex.SparsePair(ind=row, val=[1.0]*n))

    for i in range(n):
        for j in range(n):
            if i != j:
                row = []
                for k in range(n):
                    if k != i and k != j:
                        row.append(k*n+i)
                row.append(j*n+i)
                rows.append(cplex.SparsePair(ind=row, val=[1.0]*len(row)))

    for i in range(n):
        for j in range(n):
            if i != j:
                row = []
                for k in range(n):
                    if k != i and k != j:
                        row.append(i*n+k)
                row.append(i*n+j)
                rows.append(cplex.SparsePair(ind=row, val=[1.0]*len(row)))

    prob = cplex.Cplex()
    prob.objective.set_sense(prob.objective.sense.minimize)

    prob.variables.add(obj=my_obj, lb=my_lb, ub=my_ub, types=my_ctype, names=my_colnames)
    prob.linear_constraints.add(lin_expr=rows, senses=my_sense, rhs=my_rhs)

    try:
        prob.solve()
    except CplexError as exc:
        print(exc)
        return

    print("Optimal solution:", prob.solution.get_objective_value())
    x = prob.solution.get_values()
    tour = []
    for i in range(n):
        for j in range(n):
            if x[i*n+j] > 0.5:
                tour.append((i, j))
    return tour

这个算法的思路是：将每个城市看作一个节点，每个节点之间的距离为边，构建一个完全图。然后，将每个节点与它所连向的边看作变量，构建一个线性规划模型，使得每个节点都恰好被访问一次，并且路径总长度最小。利用Cplex求解这个模型，可以得到最优解。