TSP旅行商问题

求解的问题，burma.tsp里面的内容

1   16.47   96.10
2   16.47   94.44
3   20.09   92.54
4   22.39   93.37
5   25.23   97.24
6   22.00   96.05
7   20.47   97.02
8   17.20   96.29
9   16.30   97.38
10   14.05   98.12
11   16.53   97.38
12   21.52   95.59
13   19.41   97.13
14   20.09   94.55

主程序

为方便调用，已经将所有的模块整合到一个程序内。

import random
import math
import matplotlib.pyplot as plt
import re
#本程序为老师布置的旅行商问题的程序练习作业#
#制作者-zqh  联系方式：博客园-克莱比 qq:962903415#
#-------参数部分--------#
generation=[]
fitvalue=[]
# dict={
#     1:[16.47,96.10],2:[16.47,94.44],3:[20.09,92.54],
#     4:[22.39,93.37],5:[25.23,97.24],6:[22.00,96.05],
#     7:[20.47,97.02],8:[17.20,96.29],9:[16.30,97.38],
#     10:[14.05,98.12],11:[16.53,97.38],12:[21.52,95.59],
#     13:[19.41,97.13],14:[20.09,94.55]
# }
#修改部位1。此部分的字典操作应该采用文件读写的方式来完成

crossrate=4
mutationrate=1
size=30
popgeneration=400

#——————————————这里是交叉的模块——————————#
def crossOver(population):
    ret = []
    for i in range(0,int(len(population)/2)):
        cplist1=[]
        cplist2=[]
        register1=[]
        register2=[]
        crossval1=[]
        crossval2=[]
        x1=random.choice(population)
        target=0
        while target!=1:
            x2=random.choice(population)
            if x2!=x1:
                target=1
            if x2==x1:
                continue
        # print('随机抽出两个不同的母体',(x1,x2))
        while len(cplist1)<=crossrate:   #随机抽crossrate条染色体位置传给子代
            crossposition=random.randrange(0,13)
            if crossposition not in cplist1:  #保证随机取得交叉位置不重复
                cplist1.append(crossposition)
                crossval1.append(x1[crossposition])
        # print('母体1传给子代的交叉位置%s,交叉位置值%s' %(cplist1,crossval1))
        for i in range(0,14):
            if i not in cplist1:
                cplist2.append(i)
                crossval2.append(x2[i])
        # print('母体2传给子代的交叉位置%s,交叉位置值%s' %(cplist2,crossval2))
        for i1 in range(0,14):
            if i1 not in cplist1:  #单体长度14，如果是交叉位置上的就传值，如果不是就传d
                register1.append('d')
            if i1 in cplist1:
                register1.append(x1[i1])
        # print('母体1交叉位置形成子代1副本',register1)
        for i2 in x2:
            if i2 not in crossval1:  #给d更换的值不能与已经存在的值重复
                for xp in range(0,len(register1)):
                    if register1[xp]=='d': #用于定位d的位置
                        register1[xp]=i2
                        break
        # print('交叉后的子体1',register1)
        for p1 in range(0,14):
            if p1 not in cplist2:
                register2.append('d')
            if p1 in cplist2:
                register2.append(x2[p1])
        # print('母体2交叉位置形成子代1副本',register2)
        for p2 in x1:
            if p2 not in crossval2:
                for xp1 in range(0,len(register2)):
                    if register2[xp1]=='d':
                        register2[xp1]=p2
                        break
        # print('交叉后的子体2',register2)
        ret.append(register1)
        ret.append(register2)
    print('交叉完成后的种群',ret)
    return ret

#——————————————这里是变异的模块——————————#
def mutation(population):
    ret=[]
    for i in population:
        porbility=random.uniform(0,1)
        if porbility<mutationrate:
            a=random.randrange(0,14)
            b=random.randrange(0,14)
            trans=i[a]
            trans1=i[b]
            i[a]=trans1
            i[b]=trans
            ret.append(i)
        else:
            ret.append(i)
    # print('变异完成后的种群',ret)
    return ret

#——————————————————这里是产生随机种群的模块——————————#
def creatPop(popsize):
    ret1=[]
    for i in range(0,popsize):
        ret=[]
        # while(len(ret)<=13):
        #     x=random.randrange(1,15)
        #     if x not in ret:  #xulie
        #         ret.append(x)
        # ret1.append(ret)
        #根据上次说的改进，可以使用random乱序来生成
        for i1 in range(0,14):
            ret.append(i1+1)
            random.shuffle(ret)
        ret1.append(ret)
    print('生成的初代种群',ret1)
    return ret1
creatPop(2)

#——————————————————这里是将文件内的数据读取出来并放入字典的程序——————————#

dict={}

fil=open('burma14.tsp','r',encoding='utf8')   #r代表读操作，且打开时按utf8
for i in range(0,14):
    readtext=fil.readline()
    readline=re.split('[ ]',readtext)
    for i in range(0,4):
        readline.remove('')
    dict.update({int(readline[0]):[float(readline[1]),float(readline[2])]})
    print(dict)

#————————————————这里是适应度函数值的计算模块——————
def fitNess(population):
    ret=[]
    for i in population:
        # print(i)
        distance = 0
        s=0
        for xp in range(0,len(i)-1):
            # print(dict[i[s]])
            # print(dict[i[s+1]])
            distance=distance+((dict[i[s]][0]-dict[i[s+1]][0])**2+(dict[i[s]][1]-dict[i[s+1]][1])**2)**(1/2)
            # print(distance)    #每两个相邻的点位求距离
            s=s+1
        distance=distance+((dict[i[0]][0]-dict[i[13]][0])**2+(dict[i[0]][1]-dict[i[13]][1])**2)**(1/2)  #最后再求终点到起点的距离
        ret.append(distance)
    print('计算并集内个体适应度',ret)
    return ret

def popChoice(population):
    ret=[]
    ret1=[]
    dict1={}
    unionpop=[]
    for i in start:
        unionpop.append(i)
    for i1 in population:
        unionpop.append(i1)
    print('新老种群并集',unionpop)
    fitlist=fitNess(unionpop)
    for xp in range(0,len(fitlist)):
        dict1.update({fitlist[xp]:unionpop[xp]})
    print(dict1)
    fitlist.sort()
    print(fitlist)
    for xp1 in fitlist:
        ret.append(dict1[xp1])
    print(ret)
    for xp2 in range(0,size):
        if ret[xp2] not in ret1:
            ret1.append(ret[xp2])
    print(ret1)
    print('最优适应度值为',fitlist[0])
    fitvalue.append(fitlist[0])
    return ret1

a=creatPop(size)
start=a
for i in range(0,popgeneration):
    b=crossOver(start)
    c=mutation(b)
    d=popChoice(c)
    print('第%s世代,他的最优解为%s' %(i+1,d[0]))
    start=d

for i1 in range(0,popgeneration):
    generation.append(i1)

plt.plot(generation,fitvalue)
plt.ylabel('fitness value')  #为y轴加注释
plt.xlabel('generation')  #为x轴加注释
plt.show()

#本程序为老师布置的旅行商问题的程序练习作业#
#制作者-zqh  联系方式：博客园-克莱比 qq:962903415#