破圈法求最小生成樹+最小生成樹與最短路徑問題

 

 

 圖2和圖3都是樹，但是圖3是最小生成樹，他的路徑之和更小

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

代碼

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)

效果圖

 

 

 

 注意的是，最小生成樹與最短路徑是不一樣的

最小生成樹只能保證路徑之和最小而不能保證任意兩點之間路徑最小

如上圖1拓撲，最小生成樹圖2的A到D是7，2跳

最短路徑A到D是6,1跳

 

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