摘要：
x+1），（x-2，self.row=nself.column=mself.startx=xself.starty=yself.chessboard=[[0]*self.columnforrinrange（self.row+1）]self.sunx=[1，-2]self.suny=[2，-1]self.chessbboard[self.startx][self.starty]=1；
马走日字问题，在n*m的棋盘中，马只能走"日"字。马从位置(x,y)出发，把棋盘的每一格都走一次且只走一次。找出所有路径。
这个问题可以用回溯法解，每一步都有八种可能的走法，设马当前在(x,y)点，则它的可能走到：
(x+1,x+2),(x+1,x-2),(x-1,x+2),(x-1,x-2),(x+2,x+1),(x+2,x-1),(x-2,x+1),(x-2,x-1)
对每一种可能的走法试一遍，如果出界了或者已经走过了，则不用走了。试探一遍后，回溯。
python实现：
'''
horse rides sun problem
use backtracking algorithm
author:ztp
create at:2015/1/19 15:06
'''
class HorseRides:
def __init__(self, n, m, x, y):
self.row = n
self.column = m
self.startx = x
self.starty = y
self.chessboard = [[0]*self.column for r in range(self.row+1)]
self.sunx = [1, 1, 2, 2,-1,-1,-2,-2]
self.suny = [2,-2, 1,-1, 2,-2, 1,-1]
self.chessboard[self.startx][self.starty] = 1;
self.count = 0;
def check(self, x, y):
if x >= self.row or y >= self.column or x < 0 or y < 0 or self.chessboard[x][y] != 0:
return 0;
return 1;
def ride(self, x, y, step):
for i in range(8):
xx = x + self.sunx[i]
yy = y + self.suny[i]
if self.check(xx, yy) == 1:
self.chessboard[xx][yy] = step;
if step == self.row*self.column:
self.output();
else:
self.ride(xx, yy, step+1)
self.chessboard[xx][yy] = 0
def output(self):
self.count = self.count + 1
print "count = %d" % self.count
for i in range(self.row):
print self.chessboard[i]
def getCount(self):
return self.count
if __name__ == "__main__":
horseride = HorseRides(5,4,0,0)
horseride.ride(0, 0, 2)
print "total path: %d" % horseride.getCount()