标签：return network Yen python mid 最短 item range path
老规矩，先上代码
#date:2021-5-17
#author:Linuas
#b站:会武术的白猫
import copy
def Dijkstra(network,s,d):#迪杰斯特拉算法算s-d的最短路径，并返回该路径和代价
#print("Start Dijstra Path……")
path=[]#s-d的最短路径
n=len(network)#邻接矩阵维度，即节点个数
fmax=9999999
w=[[0 for i in range(n)]for j in range(n)]#邻接矩阵转化成维度矩阵，即0→max
book=[0 for i in range(n)]#是否已经是最小的标记列表
dis=[fmax for i in range(n)]#s到其他节点的最小距离
book[s-1]=1#节点编号从1开始，列表序号从0开始
midpath=[-1 for i in range(n)]#上一跳列表
u=s-1
for i in range(n):
for j in range(n):
if network[i][j]!=0:
w[i][j]=network[i][j]#0→max
else:
w[i][j]=fmax
if i==s-1 and network[i][j]!=0:#直连的节点最小距离就是network[i][j]
dis[j]=network[i][j]
for i in range(n-1):#n-1次遍历，除了s节点
min=fmax
for j in range(n):
if book[j]==0 and dis[j]<min:#如果未遍历且距离最小
min=dis[j]
u=j
book[u]=1
for v in range(n):#u直连的节点遍历一遍
if dis[v]>dis[u]+w[u][v]:
dis[v]=dis[u]+w[u][v]
midpath[v]=u+1#上一跳更新
j=d-1#j是序号
path.append(d)#因为存储的是上一跳，所以先加入目的节点d，最后倒置
while(midpath[j]!=-1):
path.append(midpath[j])
j=midpath[j]-1
path.append(s)
path.reverse()#倒置列表
#print(path)
#print(midpath)
#print(dis)
return path
def return_path_sum(network,path):
result=0
for i in range(len(path)-1):
result+=network[path[i]-1][path[i+1]-1]
return result
def add_limit(path,s):#path=[[[1,3,4,6],5],[[1,3,5,6],7],[[1,2,4,6],8]
result=[]
for item in path:
if s in item[0]:
result.append([s,item[0][item[0].index(s)+1]])
result=[list(r) for r in list(set([tuple(t) for t in result]))]#去重
return result
def return_shortest_path_with_limit(network,s,d,limit_segment,choice):#limit_segment=[[3,5],[3,4]]
mid_net=copy.deepcopy(network)
for item in limit_segment:
mid_net[item[0]-1][item[1]-1]=mid_net[item[1]-1][item[0]-1]=0
s_index=choice.index(s)
for point in choice[:s_index]:#s前面的点是禁用点
for i in range(len(mid_net)):
mid_net[point-1][i]=mid_net[i][point-1]=0
mid_path=Dijkstra(mid_net,s,d)
return mid_path
def judge_path_legal(network,path):
for i in range(len(path)-1):
if network[path[i]-1][path[i+1]-1]==0:
return False
return True
def k_shortest_path(network,s,d,k):
k_path=[]#结果列表
alter_path=[]#备选列表
kk=Dijkstra(network,s,d)
k_path.append([kk,return_path_sum(network,kk)])
while(True):
if len(k_path)==k:break
choice=k_path[-1][0]
for i in range(len(choice)-1):
limit_path=[[choice[i],choice[i+1]]]#限制选择的路径
if len(k_path)!=1:
limit_path.extend(add_limit(k_path[:-1],choice[i]))
mid_path=choice[:i]
mid_res=return_shortest_path_with_limit(network,choice[i],d,limit_path,choice)
if judge_path_legal(network,mid_res):
mid_path.extend(mid_res)
else:
continue
mid_item=[mid_path,return_path_sum(network,mid_path)]
if mid_item not in k_path and mid_item not in alter_path:
alter_path.append(mid_item)
if len(alter_path)==0:
print("总共只有{}条最短路径！".format(len(k_path)))
return k_path
alter_path.sort(key=lambda x:x[-1])
x=alter_path[0][-1]
y=len(alter_path[0][0])
u=0
for i in range(len(alter_path)):
if alter_path[i][-1]!=x:
break
if len(alter_path[i][0])<y:
y=len(alter_path[i][0])
u=i
k_path.append(alter_path[u])
alter_path.pop(u)
for item in k_path:
print(item)
return k_path
network=[[0,3,2,0,0,0],
[3,0,1,4,0,0],
[2,1,0,2,3,0],
[0,4,2,0,2,1],
[0,0,3,2,0,2],
[0,0,0,1,2,0]]
k_shortest_path(network,1,6,10)
运行结果
演示讲解
标签：return,network,Yen,python,mid,最短,item,range,path 来源： https://www.cnblogs.com/ljy1227476113/p/14779065.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。