骑士周游-马踏棋盘问题(A Knight's Journey)

看到 @大城小胖 做了个H5游戏:马踏棋盘,于是想起研究一下这个算法。

题目是这样的:国际象棋的棋盘为8*8的方格棋盘,将“马”放在任意指定的方格中,按照“马”走棋的规则将“马”进行移动。要求每个方格只能进入一次,最终使得“马”走遍棋盘64个方格。(N=8的情况)

我用python实现了这个算法,其中用到了回溯法,并用贪心法进行优化,以防递归深度太深而溢出。当找到一个解就停止递归。

不过这个解的最后一步与初始位置难以重合,而@大城小胖 的游戏中却要求马最后要回到初始位置,因此算法还要进化

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N = 6
D = 8
offsetX = [-1,1,-1,1,-2,-2,2,2]
offsetY = [-2,-2,2,2,-1,1,-1,1]
chessBoard= [[0]*N for i in xrange(N)]


class ChessNode:
def __init__(self, x, y):
self.x, self.y = x, y
self.wayout = 0

def isCorrect(self, count=0):
return self.x<N and self.x>=0 and self.y<N and self.y>=0 and (chessBoard[self.x][self.y]==0 or chessBoard[self.x][self.y]==1 and count==N*N+1)

def setWayout(self, count=0):
for direction in xrange(D):
nx, ny = self.x + offsetX[direction], self.y + offsetY[direction]
aNode = ChessNode(nx, ny)
self.wayout += 1 if aNode.isCorrect(count=count) else 0
def show(self):
print '(%s,%s)=> %s wayouts' % (self.x, self.y, self.wayout)


class HorseJump:
def success(self):
for i in xrange(N):
for j in xrange(N):
print "%2d " % chessBoard[i][j],
print
print '====================='



def greedyDFS(self, node, count):
# node.show()
# self.success()
# print count
if self.found: return
if count > N*N:
# self.found = True
self.success()
return

v = []
for direction in xrange(D):
newNode = ChessNode(node.x+offsetX[direction], node.y+offsetY[direction])
if newNode.isCorrect(count=count+1):
newNode.setWayout(count=count+1)
v.append(newNode)

sv = sorted(v, lambda x,y: x.wayout < y.wayout)
for it in sv:
chessBoard[it.x][it.y] = count
self.greedyDFS(it, count+1)
chessBoard[node.x][node.y] = 0

def greedyWalk(self, ox, oy):
self.ox, self.oy = ox, oy
node = ChessNode(ox, oy)
if not node.isCorrect():
print "illegal init x,y"
return
self.success()
chessBoard[ox][oy]=1
self.found = False
self.greedyDFS(node, count=2)


h=HorseJump()
h.greedyWalk(4,3)

参考:
http://blog.csdn.net/wangzhicheng1983/article/details/7871650
http://blog.csdn.net/cdu09/article/details/20400101


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