#!/usr/bin/env python3 # -*- coding: utf-8 -*- """ Created on Thu May 30
10:53:38 2019 @author: lg """ import pandas as pd import numpy as np
ll=[df3,df2,df1] ll1=[df1,df2] cv=[] def pp(t,df): mm=df.index f=pd.Series()
g=[] for j in mm: m1=df.loc[j] m2=m1+t m3=m2.sort_values() f[j]=m3[0]
g.append(m3.index[0]) dc=pd.DataFrame() dc['v']=f.values dc['n']=g
dc.index=f.index # print(dc) cv.append(dc) if len(ll1)>0: df=ll1.pop()
t=pp(dc['v'],df) else: return dc t1=df4['E'] h1=pp(t1,df3) for m in
range(len(cv)): xc=cv.pop() x1=xc.sort_values(by='v') print(x1['n'].values[0])
B1 C3 D2
So the shortest route is A B1 C3 D2 E, It's a little bit of a program , There may actually be multiple routes to the problem , This program only takes one

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