破圈法求最小生成树+最小生成树与最短路径问题

 

 图2和图3都是树，但是图3是最小生成树，他的路径之和更小

 

 

 

 

代码

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)

效果图

 注意的是，最小生成树与最短路径是不一样的

最小生成树只能保证路径之和最小而不能保证任意两点之间路径最小

如上图1拓扑，最小生成树图2的A到D是7，2跳

最短路径A到D是6,1跳

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