求解递归函数的路径

Question

在下面的代码中，我得到了我设置的递归函数的最佳值。
现在我想得到帮助构建它的值——我的意思是，知道在每个步骤中使用了哪个选项（或者“搭便车”（这是我的决定）），并以字符串/数组/某种方式返回。你知道吗

故事：我有6个游戏，在每一步中我需要选择是继续游戏，还是切换。每次乘坐都有一个有趣的费率，每次我去它，我想最大限度地发挥乐趣。现在我有最佳的价值，但不是我的骑乘，我去，使我到它

函数F是这里的重点。你知道吗

import math
import numpy as np

#fun rate for each ride,per loop
def funPerRide(rideNum,loopsNum):
    if rideNum==1:
        return 0

    if rideNum==2:
        return 100-np.power((4*loopsNum-10),2)

    if rideNum==3:
        return 50-(1000*np.power(np.e,(-loopsNum/2))/(5*loopsNum+20))

    if rideNum==4:
        return 4*loopsNum

    if rideNum==5:
        return 2.5*np.power(loopsNum,2)

    if rideNum==6:
        return 50*np.power(np.e,(-2*loopsNum+4))


def F(totalTime,timeLeft,rideNum,loopOnCurrRide):
    #time of line+operation of ride
    totalTimePerRide={1:0,2:40,3:15,4:20,5:23,6:11}
    #time of operation of rides
    operationTimePerRide={1:0,2:4,3:5,4:8,5:3,6:6}
     #unfeasable conditions:        
    if timeLeft<0:
        return -np.inf

    if timeLeft+loopOnCurrRide*operationTimePerRide[rideNum]>totalTime:
        return -np.inf

    if loopOnCurrRide>3:
        return -np.inf
    #edge condition
    if timeLeft == 0:
        return 0

    #fun if i stay on the ride im on right now
    staying = funPerRide(rideNum,loopOnCurrRide+1)-funPerRide(rideNum,loopOnCurrRide)+F(totalTime,timeLeft-operationTimePerRide[rideNum],rideNum,loopOnCurrRide+1)

    #calculating fun if i switch to the maximum-fun-ride, that is not the ride im currently at
    switching = -1
    whichRide=-1
    for i in range(1,7):
        if i>rideNum:
            switchOption = funPerRide(i,loopOnCurrRide)+F(totalTime,timeLeft-4.5-totalTimePerRide[i],i,1)
            if switchOption>switching:
                switching, whichRide=switchOption,i

    #calculating maximum fun between switching and staying
    maxval,maxride=max((staying,rideNum),(switching,whichRide))
    path.append(maxride)
    maxval=float(maxval)

    return float(maxval)

path = []    
print(F(120,120,1,0),path)  

Answer 1

函数F可以返回两个值：第一个最优答案，第二个最优路径作为索引列表。你知道吗

def F(totalTime, timeLeft, rideNum, loopOnCurrRide):
    # time of line+operation of ride
    totalTimePerRide = {1: 0, 2: 40, 3: 15, 4: 20, 5: 23, 6: 11}
    # time of operation of rides
    operationTimePerRide = {1: 0, 2: 4, 3: 5, 4: 8, 5: 3, 6: 6}
    if timeLeft + loopOnCurrRide * operationTimePerRide[rideNum] > totalTime:
        return -10, []

    if loopOnCurrRide > 3:
        return -10, []

    if timeLeft == 0:
        return 0, []

    staying, staying_path = F(totalTime, timeLeft - operationTimePerRide[rideNum], rideNum, loopOnCurrRide + 1)
    staying += funPerRide(rideNum, loopOnCurrRide + 1) - funPerRide(rideNum, loopOnCurrRide)
    staying_path = [-1] + staying_path

    switching = -1
    switching_path = []
    for i in range(1, 7):
        if i > rideNum:
            switchOption, switchOption_path = F(totalTime, timeLeft - 4.5 - totalTimePerRide[i], i, 1)
            switchOption += funPerRide(i, loopOnCurrRide)
            if switchOption > switching:
                switching = switchOption
                switching_path = [i] + switchOption_path

    return max((staying, staying_path), (switching, switching_path))


answer, path = F(120, 120, 1, 0)