图搜索之基于Python的迪杰斯特拉算法和弗洛伊德算法，供大家参考，具体内容如下
Djstela算法
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#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))
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FLOYD算法
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#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))
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以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持服务器之家。
原文链接：https://blog.csdn.net/sinat_33829806/article/details/54487424