破圈法求最小生成树+最小生成树与最短路径问题

import networkx as nx
import matplotlib.pyplot as plt
tupo=[[0,2,2,0,3,0],
       [2,0,5,1,4,0],
       [2,5,0,2,6,4],
       [0,1,2,0,0,3],
       [3,4,6,0,0,0],
       [0,0,4,3,0,0]]

def circle(tupo,st):
    result=[]
    def trace(path,tupo,now):
        if len(result)==1:return
        if now==st and len(path)>2:
            result.append(list(path))
            return
        for i in range(len(tupo)):
            if tupo[now][i]==0 or i in path:continue
            path.append(i)
            trace(path,tupo,i)
            path.pop()
    trace([],tupo,st)
    if len(result)!=0:
        result[0].insert(0,st)
        return result[0]
    return result

def des_cir(tupo,st):
    n=len(tupo)
    for i in range(n):
        while(1):
            res=[]
            res=circle(tupo,st)
            if len(res)==0:break
            mid_len=0
            mid_st=-1
            for j in range(len(res)-1):
                if tupo[res[j]][res[j+1]]>mid_len or (tupo[res[j]][res[j+1]]==mid_len and res[j]+res[j+1]>res[mid_st]+res[mid_st+1]):
                    mid_len=tupo[res[j]][res[j+1]]
                    mid_st=j
            tupo[res[mid_st]][res[mid_st+1]]=tupo[res[mid_st+1]][res[mid_st]]=0
    for item in tupo:
        print(item)
des_cir(tupo,0)
def draw(tupo):
    nodes=[
    ‘A‘,
    ‘B‘,
    ‘C‘,
    ‘D‘,
    ‘E‘,
    ‘F‘]
    G=nx.Graph()
    for node in nodes:
        G.add_node(node)
    edges=[]
    for i in range(len(tupo)):
        for j in range(len(tupo)):
            if tupo[i][j]!=0:
                edges.append((nodes[i],nodes[j]))
     
    r=G.add_edges_from(edges)
    nx.draw(G, with_labels=True,node_color=‘y‘,)
    plt.show()
draw(tupo)