Python实现Dijkstra与Floyd最短路径算法详解

#encoding=UTF-8
MAX=9
'''
Created on 2016年9月28日
@author: sx
'''
b=999
G=[[0,1,5,b,b,b,b,b,b],\
 [1,0,3,7,5,b,b,b,b],\
 [5,3,0,b,1,7,b,b,b],\
 [b,7,b,0,2,b,3,b,b],\
 [b,5,1,2,0,3,6,9,b],\
 [b,b,7,b,3,0,b,5,b],\
 [b,b,b,3,6,b,0,2,7],\
 [b,b,b,b,9,5,2,0,4],\
 [b,b,b,b,b,b,7,4,0]]
P=[]
D=[]
def Djstela(G,P,D):
 final=[]
 for i in range(0,len(G)):
 final.append(0)
 D.append(G[0][i])
 P.append(0)
 D[0]=0
 final[0]=1
 k=0
 for v in range(1,len(G)):
 min=999
 for w in range(0,len(G)):
  if final[w]==0 and D[w]<min:
  k=w
  min=D[w]
 final[k]=1 
 for t in range(0,len(G)):
  if min+G[k][t]<D[t]:
  D[t]=min+G[k][t]
  P[t]=k
 print("\n最短路径\n",D,"\n","\n前一个选择\n",P)
def search(x):
 print("选择的终点",x,"最短路径",D[x]) 
print("邻接矩阵\n")
for i in range(0,9):
 print(G[i])
Djstela(G, P, D)
q=input("\n请输入终点")
search(int(q))

#encoding=UTF-8
'''
Created on 2016年9月28日
@author: sx
'''
t=0
b=999
G=[[0,1,5,b,b,b,b,b,b],\
 [1,0,3,7,5,b,b,b,b],\
 [5,3,0,b,1,7,b,b,b],\
 [b,7,b,0,2,b,3,b,b],\
 [b,5,1,2,0,3,6,9,b],\
 [b,b,7,b,3,0,b,5,b],\
 [b,b,b,3,6,b,0,2,7],\
 [b,b,b,b,9,5,2,0,4],\
 [b,b,b,b,b,b,7,4,0]]
P=[[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],\
 [0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],\
 [0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0]]
D=[[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],\
 [0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],\
 [0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0]]
def Floyd(G,P,D):
 t=0
 for u in range(0,len(G)):
 for s in range(0,len(G)):
  D[u][s]=G[u][s]  
  P[u][s]=s
 for k in range(0,len(G)):
 for v in range(0,len(G)):
  for w in range(0,len(G)):
  if D[v][w]>D[v][k]+D[k][w]:
   t=t+1
   D[v][w]=D[v][k]+D[k][w]
   P[v][w]=P[v][k]  
Floyd(G, P, D)
def search(s,u):
 lenth=D[s][u]
 print("路径长度为",lenth)
 f=P[s][u]
 foot=[s,f]
 if f==u:
 print("无需规划，0步")
 while f!=u:
 f=P[f][u] 
 foot.append(f) 
 for i in range(0,len(foot)):
 if i==0:
  print("起 点____",foot[i])
 elif i==len(foot)-1:
  print("终 点____",foot[i],"步长___",G[foot[i-1]][foot[i]])
 else:
  print("第",i,"点____",foot[i],"步长___",G[foot[i-1]][foot[i]])
print("邻接矩阵")
for i in range(0,9):
 print(G[i])
s=input("请输入起点0-8\n")
u=input("请输入终点0-8\n")
Floyd(G, P, D)
search(int(s),int(u))