哎，好久沒寫博文了，其實仔細想來，時間還是蠻多的，以後還是多寫寫吧！
之前看過經典的搜索路徑方法，印象較深的也就BFS(廣度優先)，DFS(深度優先)以及A*搜索，但沒實踐過，就借八數碼問題，來通通實現遍，觀察下現象唄~~~
首先，怎麼說也得把數碼這玩意基本操作實現了唄！上代碼~
class puzzled: def __init__(self,puzzled): self.puzzled=puzzled self.__getPuzzledInfo() def __getPuzzledInfo(self): self.puzzledWid=len(self.puzzled[0]) self.puzzledHei=len(self.puzzled) self.__f1=False for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): if(self.puzzled[i][j]==0): self.zeroX=j self.zeroY=i self.__f1=True break if(self.__f1): break def printPuzzled(self): for i in range(0,len(self.puzzled)): print self.puzzled[i] print "" def isRight(self): if(self.puzzled[self.puzzledHei-1][self.puzzledWid-1]!=0): return False for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): if(i*self.puzzledWid+j+1!=self.puzzled[i][j]): if(i!=self.puzzledHei-1 or j!=self.puzzledWid-1): return False return True def move(self,dere):#0 up,1 down,2 left,3 right if(dere==0 and self.zeroY!=0): self.puzzled[self.zeroY-1][self.zeroX],self.puzzled[self.zeroY][self.zeroX] = self.puzzled[self.zeroY][self.zeroX],self.puzzled[self.zeroY-1][self.zeroX] self.zeroY-=1 return True elif(dere==1 and self.zeroY!=self.puzzledHei-1): self.puzzled[self.zeroY+1][self.zeroX],self.puzzled[self.zeroY][self.zeroX] = self.puzzled[self.zeroY][self.zeroX],self.puzzled[self.zeroY+1][self.zeroX] self.zeroY+=1 return True elif(dere==2 and self.zeroX!=0): self.puzzled[self.zeroY][self.zeroX-1],self.puzzled[self.zeroY][self.zeroX] = self.puzzled[self.zeroY][self.zeroX],self.puzzled[self.zeroY][self.zeroX-1] self.zeroX-=1 return True elif(dere==3 and self.zeroX!=self.puzzledWid-1): self.puzzled[self.zeroY][self.zeroX+1],self.puzzled[self.zeroY][self.zeroX] = self.puzzled[self.zeroY][self.zeroX],self.puzzled[self.zeroY][self.zeroX+1] self.zeroX+=1 return True return False def getAbleMove(self): a=[] if(self.zeroY!=0): a.append(0) if(self.zeroY!=self.puzzledHei-1): a.append(1) if(self.zeroX!=0): a.append(2) if(self.zeroX!=self.puzzledWid-1): a.append(3) return a def clone(self): a=copy.deepcopy(self.puzzled) return puzzled(a) def toString(self): a="" for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): a+=str(self.puzzled[i][j]) return a def isEqual(self,p): if(self.puzzled==p.puzzled): return True return False def toOneDimen(self): a=[] for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): a.append(self.puzzled[i][j]) return a def getNotInPosNum(self): t=0 for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): if(self.puzzled[i][j]!=i*self.puzzledWid+j+1): if(i==self.puzzledHei-1 and j==self.puzzledWid-1 and self.puzzled[i][j]==0): continue t+=1 return t def getNotInPosDis(self): t=0 it=0 jt=0 for i in range(0,self.puzzledHei): for j in range(0,self.puzzledWid): if(self.puzzled[i][j]!=0): it=(self.puzzled[i][j]-1)/self.puzzledWid jt=(self.puzzled[i][j]-1)%self.puzzledWid else: it=self.puzzledHei-1 jt=self.puzzledWid-1 t+=abs(it-i)+abs(jt-j) return t @staticmethod def generateRandomPuzzle(m,n,ran): tt=[] for i in range(0,m): t=[] for j in range(0,n): t.append(j+1+i*n) tt.append(t) tt[m-1][n-1]=0 a=puzzled(tt) i=0 while(i<ran): i+=1 a.move(random.randint(0,4)) return a
稍微註解一下，puzzled類表示一個數碼類，初始化利用
a=puzzled( [1,2,3], [4,5,6], [7,8,0])
其中呢，0表示空格位置，上面初始化的便是一個正確的，未被打亂的位置~
其他的成員函數，看名稱就很好理解了唄~
ok，基礎打好了，接下來就該上節點類了：
class node: def __init__(self,p): self.puzzled=p self.childList=[] self.father=None def addChild(self,child): self.childList.append(child) child.setFather(self) def getChildList(self): return self.childList def setFather(self,fa): self.father=fa def displayToRootNode(self): t=self tt=0 while(True): tt+=1 t.puzzled.printPuzzled() t=t.father if(t==None): break print "it need "+str(tt)+ " steps!" def getFn(self): fn=self.getGn()+self.getHn() #A* #fn=self.getHn() #貪婪 return fn def getHn(self): Hn=self.puzzled.getNotInPosDis() return Hn def getGn(self): gn=0 t=self.father while(t!=None): gn+=1 t=t.father return gn
對於節點類吧，也還是很好理解的，初始化方法
a=node( puzzled([1,2,3], [4,5,6], [7,8,0]) )
基礎都搭好了，重點人物該閃亮登場了唄~
class seartchTree: def __init__(self,root): self.root=root def __search2(self,hlist,m): #二分查找，經典算法，從大到小，返回位置 #若未查找到，則返回應該插入的位置 low = 0 high = len(hlist) - 1 mid=-1 while(low <= high): mid = (low + high)/2 midval = hlist[mid] if midval > m: low = mid + 1 elif midval < m: high = mid - 1 else: return (True,mid) return (False,mid) def __sortInsert(self,hlist,m):#對於一個從大到小的序列， #插入一個數，仍保持從大到小 t=self.__search2(hlist,m) if(t[1]==-1): hlist.append(m) return 0 if(m<hlist[t[1]]): hlist.insert(t[1]+1, m) return t[1]+1 else: hlist.insert(t[1], m) return t[1] def breadthFirstSearch(self):#廣度優先搜索 numTree=NumTree.NumTree() numTree.insert(self.root.puzzled.toOneDimen()) t=[self.root] flag=True generation=0 while(flag): print "it's the "+str(generation)+" genneration now,the total num of items is "+str(len(t)) tb=[] for i in t: if(i.puzzled.isRight()==True): i.displayToRootNode() flag=False break else: for j in i.puzzled.getAbleMove(): tt=i.puzzled.clone() tt.move(j) a=node(tt) if(numTree.searchAndInsert(a.puzzled.toOneDimen())==False): i.addChild(a) tb.append(a) t=tb generation+=1 def depthFirstSearch(self):#深度優先搜索 numTree=NumTree.NumTree() numTree.insert(self.root.puzzled.toOneDimen()) t=self.root flag=True gen=0 while(flag): bran=0 print "genneration: "+str(gen) if(t.puzzled.isRight()==True): t.displayToRootNode() flag=False break else: f1=True for j in t.puzzled.getAbleMove(): tt=t.puzzled.clone() tt.move(j) a=node(tt) if(numTree.searchAndInsert(a.puzzled.toOneDimen())==False): t.addChild(a) t=a f1=False gen+=1 break if(f1==True): t=t.father gen-=1 def AStarSearch(self):#A* numTree=NumTree.NumTree() numTree.insert(self.root.puzzled.toOneDimen()) leaves=[self.root] leavesFn=[0] while True: t=leaves.pop() #open表 print leavesFn.pop() if(t.puzzled.isRight()==True): t.displayToRootNode() break for i in t.puzzled.getAbleMove(): tt=t.puzzled.clone() tt.move(i) a=node(tt) if(numTree.searchAndInsert(a.puzzled.toOneDimen())==False):#close表 t.addChild(a) fnS=self.__sortInsert(leavesFn,a.getFn()) leaves.insert(fnS, a)
注意到裏面存在一個 NumTree，這個是個啥玩意了，這個吧，其實是一個closed表，啥是Closed表呢？
就是呀，在搜索的時候，得記錄已經擴展的節點，不然的話很有可能成爲不完備的搜索了（對於深度搜索），而且已經擴展的狀態，也對於數碼問題沒必要在擴展一次，那麼怎麼來實現記錄已經擴展的節點呢？
一個最簡單的方法就是建立一個鏈表，將擴展的節點加進去，但這樣存在一個問題，就是由於每次在擴展節點時都得遍歷closed表，查找是否存在同樣的節點已經被擴展，這樣，複雜度就簡直太高了，完全不行唄~
那麼咋辦呢？從上面代碼注意到，node對象裏面有一個Puzzled成員，實際上puzzled對象主要就由一個二維數組構成唄~這時候呀，我就把這個二維數組變爲一維唄
[1,2,3], [4,5,6], -> [1,2,3,4,5,6,7,8,0] [7,8,0]
然後呢，利用trie樹存儲該數組，這樣一來，查找添加一步到位！
也就是上面代碼中的這個啦~
numTree.searchAndInsert(a.puzzled.toOneDimen())
貼上numtree代碼：
class NumTree: def __init__(self): self.root = Node() def insert(self, key): # key is of type string # key should be a low-case string, this must be checked here! node = self.root for char in key: if char not in node.children: child = Node() node.children[char] = child node = child else: node = node.children[char] node.value = key def search(self, key): node = self.root for char in key: if char not in node.children: return None else: node = node.children[char] return node.value def display_node(self, node): if (node.value != None): print node.value for char in node.children.keys(): if char in node.children: self.display_node(node.children[char]) return def display(self): self.display_node(self.root) def searchAndInsert(self,m): if(self.search(m)==None): self.insert(m) return False return True """test trie = NumTree() print trie.searchAndInsert([1,2,3,4,5,6,7,8]) print trie.searchAndInsert([1,2,3,4,5,6,7,8,9]) trie.display() """
好了，這樣再看searchTree的代碼就比較清晰了唄，
哎，好吧，我承認這等低劣之作也只有我等渣渣才能寫出，其實再看時，可以發現很多地方都可以優化的，比如二維轉一維，這個其實花費的不少時間，其實可以在內部以一維形式存在的~