图搜索之基于Python的迪杰斯特拉算法和弗洛伊德算法

Djstela算法

#encoding=UTF-8MAX=9'''Created on 2016年9月28日@author: sx'''b=999G=[[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))

FLOYD算法

#encoding=UTF-8'''Created on 2016年9月28日@author: sx'''t=0b=999G=[[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))