def bigSup_cycle_distance(bigSup):
list5 = [] # 运来存放每个超级顾客内部的最长距离
for j in range(len(bigSup)):
list1 = []
for l in range(len(bigSup[j]) - 1):
for m in range(l, len(bigSup[j])):
list1.append(cf.Point3Distance(bigSup[j][l], bigSup[j][m]))
list1.sort()
list1.reverse()
if len(list1) == 0: # 由于超级顾客在形成中产生了空超级顾客，所以有此判断；若不会形成空超级顾客，则可删除
list5.append(0)
else:
list5.append(list1[0])
cycle_distance = []
for j in range(len(list5)):
cycle_distance.append(0.71 *
((list5[j]**2 / 2 * len(bigSup[j]))**(1 / 2)))
return cycle_distance
def get_big(i, table, bigSup, bigSupDistance, point3, bigSup_sizes):
totalmoney = 0
totalopen = 0
m1 = 0
m2 = 0
m3 = 0
m4 = 0
m5 = 0
allsizesize = 0
x_location = []
for j in range(cf.numOfPoint2):
if any(table[j][k] == 1 for k in range(len(bigSup))):
x_location.append(j)
totalopen = len(x_location)
totalmoney = totalopen * cf.fL
m1 += totalopen * cf.fL
for j in x_location:
totalmoney += 2 * cf.point_12_distance[0][j] * 10
m2 += 2 * cf.point_12_distance[0][j] * 10 # 一级到二级的运输成本
final_size = 0
for j in range(cf.numOfPoint3):
for k in range(5 - i):
for l in range(2):
final_size += point3[j].P[5 - k][l]
totalmoney += final_size * cf.moneyofProcessingExpress[i][3]
m3 += final_size * cf.moneyofProcessingExpress[i][3] # 二级点的处理成本
list5 = [] # 运来存放每个超级顾客内部的最长距离
for j in range(len(bigSup)):
list1 = []
for l in range(len(bigSup[j]) - 1):
for m in range(l, len(bigSup[j])):
list1.append([cf.Point3Distance(l, m), l, m])
list1.sort()
list1.reverse()
if len(list1) == 0: # 由于超级顾客在形成中产生了空超级顾客，所以有此判断；若不会形成空超级顾客，则可删除
list5.append(0)
else:
list5.append(list1[0][0])
cycle_distance = []
for j in range(len(list5)):
cycle_distance.append((list5[j]**2 / 2 * len(bigSup[j]))**(1 / 2))
tt = 0 # 记录空超级顾客的数量
distance = 0
for j in range(len(cycle_distance)):
if cycle_distance[j] == 0:
tt += 1
for j in x_location:
for k in range(len(bigSup)):
if table[j][k] == 1 and cycle_distance[k] != 0:
distance += 2 * bigSupDistance[k][j]
totaldistance = distance + sum(cycle_distance)
totalmoney += 5 * totaldistance
m4 += 5 * totaldistance
'''for j in x_location:
for k in range(len(bigSup)):
if table[j][k] == 1:
if list5[k]==0: # 由于超级顾客在形成中产生了空超级顾客，所以有此判断；若不会形成空超级顾客，则可删除
totaldistance=0
else:
totaldistance = ((list5[k] ** 2) / 2) * len(bigSup[k]) ** (1 / 2) + 2 * 0.71 * bigSupDistance[k][j]
print('totaldistance=',totaldistance)
totalmoney += 5 * totaldistance
m4 += 5 * totaldistance # 二级点到超级顾客和超级顾客内部的运输成本'''
'''for j in x_location:
for k in range(len(bigSup)):
list1 = [] # 用来存放已经开放的中转中心所服务的三级点之间的距离和编号
if table[j][k] == 1:
for l in range(len(bigSup[k])):
for m in range(l+1,len(bigSup[k])):
list1.append([cf.Point32Distance(l,m),l,m])
list1.sort()
list1.reverse()
if len(list1)==0:
totaldistance=((((0/(2**(1/2)))**2)*len(bigSup[k]))**(1/2))*0.71+2*bigSupDistance[k][j]
else:
totaldistance=((((list1[0][0]/(2**(1/2)))**2)*len(bigSup[k]))**(1/2))*0.71+2*bigSupDistance[k][j]
totalmoney+=5*totaldistance
m4+=5*totaldistance # 二级点到超级顾客和超级顾客内部的运输成本'''
final_size = 0
for j in range(cf.numOfPoint3):
for k in range(5 - i):
for l in range(2):
final_size += point3[j].P[5 - k][l]
totalmoney += final_size * cf.moneyofProcessingExpress[i][2]
m5 += final_size * cf.moneyofProcessingExpress[i][2] # 三级点的处理成本
return totalmoney, m1, m2 + m4, m3 + m5
def superCustomer(i, point3): # 生成两类超级顾客，为列表，每一个元素代表该超级顾客的顾客点的组成
# 用来存放超级顾客信息
smallSup = []
bigSup = []
# 根据该划分值求三级网点普通和大件的量
for j in range(cf.numOfPoint3):
# 求普通件的量
sumnum = 0
for t in range(i + 1):
sumnum = sumnum + point3[j].P[t][0]
point3[j].small = sumnum
# 求大件的量
sumnum = 0
for t in range(5 - i):
sumnum = sumnum + point3[j].P[5 - t][0]
point3[j].big = sumnum
# print("i=%d small=%d big=%d"%(j,point3[j].small,point3[j].big))
# 生成普通件超级顾客，用V2进行运输
pointleft = np.zeros(cf.numOfPoint3) # 点被使用与否的标志，0表示未划分，1表示划分
car = cf.car_loading[1][2 * i]
##开始模拟生成
while (isAllUsed(pointleft) == 0):
oneSuperCustomer = [] # 一个超级顾客成员
wholeExpressNum = 0 # 一个超级顾客总运输量
wholeTime = 0 # 一个超级顾客花费总时间
lastPoint = -1 # 上一个待加入的点
while (car > wholeExpressNum and wholeTime < 100):
# 第一个点，顺序选择未加入的点加入
if wholeTime == 0:
nowPoint = -1 # 待加入的点
for j in range(cf.numOfPoint3):
if pointleft[j] == 0:
nowPoint = j
stayTime = np.random.randint(15, 20) # 每个点花费的时间
oneSuperCustomer.append(nowPoint)
pointleft[nowPoint] = 1
wholeExpressNum = wholeExpressNum + point3[
nowPoint].small
wholeTime = wholeTime + stayTime
lastPoint = nowPoint
break
# 除第一个点按照距离最近的点加入
else:
distance = 1000 # 定义一个超长距离
nowPoint = -1 # 定义待加入的点
for j in range(cf.numOfPoint3):
if cf.Point3Distance(
lastPoint, j
) < distance and lastPoint != j and pointleft[j] == 0:
distance = cf.Point3Distance(lastPoint, j)
nowPoint = j
# 判断能不能加入这个点
stayTime = np.random.randint(15, 20) # 点花费的时间
if wholeTime + stayTime > 100 or wholeExpressNum + point3[
nowPoint].small > car or nowPoint == -1:
break
else:
oneSuperCustomer.append(nowPoint)
pointleft[nowPoint] = 1
wholeExpressNum = wholeExpressNum + point3[nowPoint].small
wholeTime = wholeTime + stayTime
lastPoint = nowPoint
smallSup.append(oneSuperCustomer)
# 生成大件超级顾客，用V3进行运输
pointleft = np.zeros(cf.numOfPoint3) # 点被使用与否的标志，0表示未划分，1表示划分
car = cf.car_loading[2][2 * i + 1]
##开始模拟生成
while (isAllUsed(pointleft) == 0):
oneSuperCustomer = [] # 一个超级顾客成员
wholeExpressNum = 0 # 一个超级顾客总运输量
wholeTime = 0 # 一个超级顾客花费总时间
lastPoint = -1 # 上一个待加入的点
while (car > wholeExpressNum and wholeTime < 180):
# 第一个点，顺序选择未加入的点加入
if wholeTime == 0:
nowPoint = -1 # 待加入的点
for j in range(cf.numOfPoint3):
if pointleft[j] == 0:
nowPoint = j
stayTime = np.random.randint(15, 20) # 每个点花费的时间
oneSuperCustomer.append(nowPoint)
pointleft[nowPoint] = 1
wholeExpressNum = wholeExpressNum + point3[nowPoint].big
wholeTime = wholeTime + stayTime
lastPoint = nowPoint
break
# 除第一个点按照距离最近的点加入
else:
distance = 1000 # 定义一个超长距离
nowPoint = -1 # 定义待加入的点
for j in range(cf.numOfPoint3):
if cf.Point3Distance(
lastPoint, j
) < distance and lastPoint != j and pointleft[j] == 0:
distance = cf.Point3Distance(lastPoint, j)
nowPoint = j
# 判断能不能加入这个点
stayTime = np.random.randint(15, 16) # 点花费的时间
if wholeTime + stayTime > 180 or wholeExpressNum + point3[
nowPoint].big > car or nowPoint == -1:
break
else:
oneSuperCustomer.append(nowPoint)
pointleft[nowPoint] = 1
wholeExpressNum = wholeExpressNum + point3[nowPoint].big
wholeTime = wholeTime + stayTime
lastPoint = nowPoint
bigSup.append(oneSuperCustomer)
# 返回
return smallSup, bigSup