庞果网（最小操作数）

从庞果网上，看到一题，是求最小操作数的，具体题目如下：

现用python代码实现如下：

  1 #!usr/bin/env python
  2 #coding:utf-8
  3 
  4 def compareWord(A,B):
  5     wordLen = len(A)
  6     diff = 0
  7     for i in range(wordLen):
  8         if A[i] != B[i]:
  9             diff += 1
 10     return diff
 11 
 12 
 13 def findpath(startD, endD, Dict, dictMatrix):
 14     paths = dict()
 15     minDis = min(min([dictMatrix[Dict.index(si)][Dict.index(ei)] for si in startD] for ei in endD))
 16 
 17     if minDis == 0:
 18         for si in startD:
 19             for ei in endD:
 20                 bdInd = Dict.index(si)
 21                 edInd = Dict.index(ei)
 22                 if dictMatrix[bdInd][edInd] == 0:
 23                     paths[si] = ei
 24         return paths
 25     elif minDis == 1:
 26         for si in startD:
 27             for ei in endD:
 28                 bdInd = Dict.index(si)
 29                 edInd = Dict.index(ei)
 30                 if dictMatrix[bdInd][edInd] == 1:
 31                     if si not in paths:
 32                         paths.setdefault(si, {})
 33                     paths[si] = ei
 34         return paths
 35     else:
 36         for si in startD:
 37             bdInd = Dict.index(si)
 38             temp = list()
 39             tempArray = dictMatrix[bdInd]
 40             for t in range(len(tempArray)):
 41                 if tempArray[t] == 1:
 42                     temp.append(Dict[t])
 43             path = findpath(temp, endD, Dict, dictMatrix)
 44             paths[si] = path
 45     return paths
 46 
 47 
 48 def translate(A, B , Dict):
 49     diff = compareWord(A,B)
 50     if diff == 0:
 51         return []
 52     paths = list()
 53     #找到入口
 54     startD = list()
 55     #出口
 56     endD = list()
 57     for di in Dict:
 58         diff = compareWord(A, di)
 59         if diff == 1:
 60             startD.append(di)
 61         elif diff == 0:
 62             startD=[di]
 63             break
 64     for di in Dict:
 65         diff = compareWord(B, di)
 66         if diff == 1:
 67             endD.append(di)
 68         elif diff == 0:
 69             endD=[di]
 70             break
 71     #构建单词集合的矩阵
 72     dictMatrix = list()
 73     for di in Dict:
 74         temp = list()
 75         for dj in Dict:
 76             if di == dj:
 77                 temp.append(0)
 78             else:
 79                 temp.append(compareWord(di,dj))
 80         dictMatrix.append(temp)
 81 
 82     for di in dictMatrix:
 83         print di
 84 
 85     #寻找startD到endD的最短路径
 86     print 'startD:', A, startD
 87     print 'endD:', B, endD
 88     paths = findpath(startD, endD, Dict, dictMatrix)
 89     print paths
 90     pathSt = [A]
 91     pathSt = printDict(paths, pathSt)
 92     pathSt = [pi+'->'+B for pi in pathSt]
 93     return  pathSt
 94 
 95 def printDict(paths, pathSt):
 96     nextpath = paths.keys()
 97     pathSt = [ps+'->' for ps in pathSt]
 98     pathSt = pathSt*len(nextpath)
 99     for ni in range(len(nextpath)):
100         pathSt[ni] += nextpath[ni]
101         if type(paths[nextpath[ni]]) is not type(dict()):
102             if not pathSt[ni].endswith(paths[nextpath[ni]]):
103                 pathSt[ni] += '->'+paths[nextpath[ni]]
104             #return pathSt
105         else:
106             pathSt = printDict(paths[nextpath[ni]], pathSt)
107     return pathSt
108 
109 if __name__=='__main__':
110     A = 'hog'
111     B = 'hog'
112     Dict = ['hot', 'dot', 'dog', 'lot', 'log','tog','tit']
113     paths = translate(A,B, Dict)
114     print paths

 

posted @ 2013-10-30 11:10  搞点儿算法  阅读(343)  评论(0)  编辑  收藏  举报