格式化时间

365 Or 366？

一马当先

dx=(-1,1,2,2,1,-1,-2,-2)dy=(2,2,1,-1,-2,-2,-1,1)find=Falseque=[]que.append((n,0,0))v = [[1 for i in range(m+2)] for j in range(n+2)]v[n][0]=0while len(que)&gt;0: t=que[0] que.pop(0) x=t[0] y=t[1] step=t[2] if x==0 and y== m: print(step) find=True break for i in range(8): tx=x+dx[i] ty=y+dy[i] if tx&lt;0 or tx&gt;n or ty&lt;0 or ty&gt;m: continue if v[tx][ty]: que.append((tx,ty,step+1)) v[tx][ty]=0if not find: print(-1)

BFS求最短路径

<a href="http://www.gznb.xyz/posts/pythonTip-24-一马当先/">题解： http://www.gznb.xyz/posts/pythonTip-24-一马当先/</a>

pythonTip-24-一马当先

# 生成马的日字走法 def dimension(m = [], x = 1, y = 2, l = [0, 0]): for i in [[x, y], [y, x]]: for j in [-1, 1]: for k in [-1, 1]: m.append([(l[0] + i[0] * j), (l[1] + i[1] * k)]) return m # 去重 def formate_list(org_l = [[]], formate_l = [[]]): for i in org_l: if i not in formate_l: formate_l.append(i) return formate_l # 初始化参数 l = [[0, 0]] c = -1 # 每一步拓展列表 def step(l = [[0, 0]], d = dimension(), a = n, b = m): for i in l.copy(): for j in d: # 限定在格子范围里 if 0 &lt;= int(i[0]) + int(j[0]) &lt;= a and 0 &lt;= int(i[1]) + int(j[1]) &lt;= b: l.append([(i[0] + j[0]), (i[1] + j[1])]) formate_list(l) l.remove(i) return l # 运算 for i in range(m * n): l = step(l) if [n, m] in l or [m, n] in l: c = i + 1 break print(c)

初学Python，解得很吃力，新手共勉

<pre><code>import queue n = 1 m = 3 max_num = 100000000 L = [[max_num] * (m + 1) for i in range(n + 1)] L[0][0] = 0 book = [[False] * (m + 1) for i in range(n + 1)] book[0][0] = True q = queue.Queue() q.put((0, 0)) direction = [[1, 2], [1, -2], [-1, 2], [-1, -2], [2, 1], [2, -1], [-2, 1], [-2, -1]] while not q.empty(): x, y = q.get() for dire in direction: xx = x + dire[0] yy = y + dire[1] if 0 &lt;= xx &lt;= n and 0 &lt;= yy &lt;= m and not book[xx][yy]: L[xx][yy] = min(L[xx][yy], 1 + L[x][y]) q.put((xx, yy)) book[xx][yy] = True print(L[n][m] if L[n][m] != max_num else -1)</code></pre>

我也是想到8个方向，但是写不来，从你这学习下，点个赞

8个方向一个个试吧，感觉写复杂了

<pre><code>def horse(n, m): def func(n, m):    a = min(n, m)     b = max(n, m) if a == 1 and b != 2: print -1 elif (a + b) % 3 == 0: print (a + b) // 3 elif (a + b) % 3 == 1: print (a+b)//3 + 3 else: print (a+b)//3 + 2 return return func(n, m) horse(n, m)</code></pre>

n = 1  m = 6    一行 六列，输出 -1   其实用三步可以走到右上角

算是找规律吧

i=0 if (n==1 or m==1) and (m*n)%4!=2: print -1 else: if m&gt;n:         m,n=n,m while m&gt;0 and n&gt;0:         m,n=m-1,n-2 if n&lt;0:             n,i=n+3,i-1         i=i+1 print(i+((m+n)+(m+n)%4)/2 if (m+n)%2==0 else i+2+((m+n)+(m+n)%4)/2)

直接从程序框复制过来的，可能再复制回去的时候缩进不对，我提交是能通过的

感谢分享
实测无法通过，还有格式改起来很麻烦
希望能用四空格或tab缩进

看到其他人的解法太复杂了，给个简单的

只需要注意一点 Python2 和Python3 对queue的处理是有差别的，名字不同，干脆直接用list写模拟一个马的行为 map是最短距离 used记录已经使用的点，让这个马在点阵里面跳，先进先出以保证第一个跳到终点的马是最快的<pre class="prettyprint lang-py linenums">map = [[100000000 for x in range(m+1)] for y in range(n+1)]
used = [[0 for x in range(m+1)] for y in range(n+1)] 
map[0][0] = 0
dx=[1,-1,1,-1,2,-2,2,-2]
dy=[2,-2,-2,2,1,-1,-1,1]
horses = []
horses.append([0,0])
while(horses):
 h = horses[0]
 del horses[0]
 if h[0]==n and h[1]==m:
 break
 for i in range(8):
 nx=h[0]+dx[i]
 ny=h[1]+dy[i]
 if nx&gt;=0 and nx&lt;=n and ny&gt;=0 and ny&lt;=m and used[nx][ny]==0:
 map[nx][ny]=map[h[0]][h[1]]+1
 used[nx][ny]=1
 horses.append([nx,ny])
if map[n][m]!=100000000:
 print(map[n][m])
else:
 print(-1)</pre>

系统显示复杂度过高

见&nbsp;https://blog.csdn.net/marxoffice/article/details/97501536

简单bfs

详情请跳转：<a href="https://blog.csdn.net/Herishwater/article/details/94386635">https://blog.csdn.net/Herishwater/article/details/94386635</a>

广度优先算法（BFS）讲解

class Hourse: &nbsp; &nbsp; allSteps = [] &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; def __init__(self,w,h): &nbsp; &nbsp; &nbsp; &nbsp; self.w = w &nbsp; &nbsp; &nbsp; &nbsp; self.h = h &nbsp; &nbsp; &nbsp; &nbsp; self.x = 0 &nbsp; &nbsp; &nbsp; &nbsp; self.y = 0 &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; def success(self):#判断是否走到终点 &nbsp; &nbsp; &nbsp; &nbsp; if self.x == self.w and self.y == self.h:#要改终点，就该这里就行了 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return True &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; def run(self,x,y,steps):##平移x,y，并判断是否出界或已走过。如果是，则此步试探成功，否则失败 &nbsp; &nbsp; &nbsp; &nbsp; x2 = self.x+x &nbsp; &nbsp; &nbsp; &nbsp; y2 = self.y+y &nbsp; &nbsp; &nbsp; &nbsp; if x2 &lt; 0 or x2 &gt; self.w or y2 &lt; 0 or y2 &gt; self.h or steps.count((x2,y2)) &gt; 0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False; &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.x = x2 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.y = y2 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps.append((x2,y2)) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return True &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; def attemp(self,steps): #渗搜递归递归 &nbsp; &nbsp; &nbsp; &nbsp; if self.success():#成功则把此路径放到类变量列表中,回退一步，继续上一步的探路 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Hourse.allSteps.append(steps[::]) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lastStep = steps.pop() &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.run(-lastStep[0],-lastStep[1],[]) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(1,2,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(1,-2,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(2,1,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(2,-1,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(-1,2,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(-1,-2,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(-2,1,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; steps1 = steps[::] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if self.run(-2,-1,steps1): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; self.attemp(steps1) &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; lastStep = steps.pop()#失败也回退一步，继续上一步的探路 &nbsp; &nbsp; &nbsp; &nbsp; self.run(-lastStep[0],-lastStep[1],[]) &nbsp; &nbsp; &nbsp; &nbsp; return &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; @classmethod &nbsp; &nbsp; def getsp(cls):#获取成功路径列表中第一个最短的路径 &nbsp; &nbsp; &nbsp; &nbsp; if len(cls.allSteps)==0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return [] &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return sorted(cls.allSteps,lambda x,y:cmp(len(x),len(y)))[0] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; hourse = Hourse(m,n)#宽高 hourse.attemp([(0,0)])#初始化起点 #print(Hourse.getsp())#可打印路径，检查结果 pathLength = len(Hourse.getsp())-1;#第一步起点要去掉 print(-1 if pathLength == 0 else pathLength)

面对对象的深搜，可求任意两点的最短路径和长度

t=(n,m) s=(0,0) q=set() for i in range(-1,2,2): &nbsp; &nbsp; for j in range (-2,3,4): &nbsp; &nbsp; &nbsp; &nbsp; q=q|set([(i,j),(j,i)]) p=set([s]) def sp(p,q,n,m): &nbsp; &nbsp; r=p &nbsp; &nbsp; for x in p: &nbsp; &nbsp; &nbsp; &nbsp; for y in q: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if x[0]+y[0]&gt;=0 and x[1]+y[1]&gt;=0 and x[1]+y[1]&lt;=m and x[0]+y[0]&lt;=n: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; z=(x[0]+y[0],x[1]+y[1]) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if z in p: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; r=r-set([z]) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; r=r|set([z]) &nbsp; &nbsp; return r i=0 while t not in p and i&gt;=0: &nbsp; &nbsp; p=sp(p,q,n,m) &nbsp; &nbsp; i+=1 &nbsp; &nbsp; if i&gt;=3*max(m,n): &nbsp; &nbsp; &nbsp; &nbsp; i=-1 print i

穷举每步能到的点集，去重做的不好，-1也随便找了个边界

<pre><code>plus=m+n if plus%3 == 0: a=int(plus/3) if(m &lt;= plus-a and n &lt;= plus-a) and (m &gt;= a and n &gt;= a): print(str(a)) elif (m==1 or n==1) and plus%3 != 0 : if m==1: if((n/2)%2==1): print(int(n/2)) elif n==1: if((m/2)%2==1): print(int(m/2)) else: print(&quot;-1&quot;) 原来的代码少了两种情况，加上了之后还是执行能通过但提交过不了</code></pre>

<pre><code>plus=m+n if plus%3 == 0:     a=int(plus/3) if(m &lt;= plus-a and n &lt;= plus-a) and (m &gt;= a and n &gt;= a): print(str(a)) else: print(&quot;-1&quot;)</code></pre>

执行代码过了但提交代码没过是什么鬼

#row_step*2 + col_step = m #row_step + col_step*2 = n

from sympy import * import re n = 5 m = 4 row_step = Symbol(&#x27;row_step&#x27;) col_step = Symbol(&#x27;low_step&#x27;) result = solve([row_step*2 + col_step-m,row_step + col_step*2-n],[row_step,col_step]) result_row_step = result[row_step] result_col_step = result[col_step] if re.search(&#x27;/&#x27;, str(result_col_step)) and re.search(&#x27;/&#x27;, str(result_row_step)): print(&#x27;-1&#x27;) else: print(result_col_step+result_row_step)

sympy模块能正确计算，但是找不到模块

<pre style="background-color:#2b2b2b;color:#a9b7c6;font-family:'宋体';font-size:12.0pt;">import numpy as np

def solveyima(n,m):
 a = np.array([[2,1],[1,2]])
 bl = [n,m]
 b = np.array(bl)
 x = np.linalg.solve(a,b)
 if x[0] &gt; 0 and x[1]&gt;0 and x[0]%1 == 0 and x[1]%1 == 0:
 return int(x[0] + x[1])
 else:
 return -1</pre>

用矩阵思想简单

def arrival(x,y,m,n): &nbsp;&nbsp;&nbsp; if x==m and y == n: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return 1 &nbsp;&nbsp;&nbsp; else:return 0 def run(m,n,b,a=[],x=1,y=1,road=([1,2],[2,1],[1,-2],[2,-1])): &nbsp;&nbsp;&nbsp; try: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; for i in range(4): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [i1, i2] = road[i] &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if x + i1 &lt;= m and x + i1 &gt; 0 and y + i2 &lt;= n and y + i2 &gt; 0: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x=x+i1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y=y+i2 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a.append(i+1) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if arrival(x, y, m, n): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; # print('至少需要走的步数为：%d，\n具体走法如下：' %len(a),a) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; b.append(len(a)) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; run(m,n,x=x,y=y,b=b,a=a) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; x = x - i1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; y = y -i2 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a.pop() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; # break &nbsp;&nbsp;&nbsp; except RuntimeError:print('棋盘过大') def start(m,n): &nbsp;&nbsp;&nbsp; b=[] &nbsp;&nbsp;&nbsp; run(m, n, b) &nbsp;&nbsp;&nbsp; if&nbsp; b: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print('至少需要走的步数为：%d'%min(b)) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return min(b) &nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print('没有结果返回') &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return -1 start(m,n)

感觉这个简单一些，但是好像通不过，说变量不同

if n&gt;m:m,n=n,m def move(L,count,mode): &nbsp;&nbsp;&nbsp; if mode==&quot;N&quot;:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #到顶或者顶减一 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if L[1]==n-1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[0] += 2&nbsp; # 靠n右上移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[1] += 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[0] += 2&nbsp; # 右下移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[1] -= 1 &nbsp;&nbsp;&nbsp; elif mode==&quot;M&quot;: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if L[0]==m-1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[1] += 2&nbsp; # 靠m右上移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[0] += 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[1] += 2&nbsp; # 左上移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L[0] -= 1 &nbsp;&nbsp;&nbsp; count[0] += 1 def NewWay(L,count): &nbsp;&nbsp;&nbsp; flag=0 &nbsp;&nbsp;&nbsp; if L[1]==n-1 or L[1]==n:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #到上面了，就向右移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; while flag!=1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; move(L,count,&quot;N&quot;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if isEnd(L): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; flag=1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; break &nbsp;&nbsp;&nbsp; if L[0]==m-1 or L[0]==m:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #到右边了，就向上移动 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; while flag!=1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; move(L,count,&quot;M&quot;) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if isEnd(L): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; flag=1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; break &nbsp;&nbsp;&nbsp; return L def isEnd(L):&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #判断是否走到特殊位置，即与目标位置横纵距离都小于2 &nbsp;&nbsp;&nbsp; if L[0]==m and n-L[1]&lt;2: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return True &nbsp;&nbsp;&nbsp; elif L[1]==n and m-L[0]&lt;2: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return True &nbsp;&nbsp;&nbsp; elif L[1]==n-1 and L[0]==m-1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return True &nbsp;&nbsp;&nbsp; else:return False def Position(L): &nbsp;&nbsp;&nbsp; if L[0]==m-1 and L[1]==n-1:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #终点左下一格 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return 2 &nbsp;&nbsp;&nbsp; elif L[0]==m and L[1]==n:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; #刚好终点 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return 0 &nbsp;&nbsp;&nbsp; elif L[0]==m-1 or L[1]==n-1:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; # &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return 1 &nbsp;&nbsp;&nbsp; else:return False def main(m,n): &nbsp;&nbsp;&nbsp; count = [0]&nbsp; # 这里要用list是为了传参数引用 &nbsp;&nbsp;&nbsp; O1 = [0, 0] &nbsp;&nbsp;&nbsp; while (O1[0] + 2 &lt;= m and O1[1] + 1 &lt;= n): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; O1[0] += 2 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; O1[1] += 1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; count[0] += 1 &nbsp;&nbsp;&nbsp; if n &lt;= 1: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if (m - 2) % 4 != 0: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print(-1) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else:print(int(m/2)) &nbsp;&nbsp;&nbsp; elif m==n==2: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print(4) &nbsp;&nbsp;&nbsp; elif Position(O1): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print(count[0] + Position(O1)) &nbsp;&nbsp;&nbsp; else: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if not isEnd(O1): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; print(Position(NewWay(O1, count)) + count[0])&nbsp; # 这里顺序不能错 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; else:print(count[0]) main(m,n)

测试通过，代码没有优化，供各位同学参考

可不可以看成从(0,0)这个坐标出发到达(n,m)这个终点,其中可以走的步骤有(1,2)(2,1)(-1,2)(2,-1)(-2,1)(1,-2)(-1,-2)(-2,-1)这8种 然后通过每个不走的系数调整,最终达到终点

n = -1 m = -2 a = [1,2]b = [2,1] n = abs(n) m = abs(m)new = [n,m]maxnumber = max(m,n) result = -1 for i in range(maxnumber+1): &nbsp; &nbsp; for j in range(maxnumber+1): &nbsp; &nbsp; &nbsp; &nbsp; if ((i * a[0] + j * b[0] == new[0])&amp;(i * a[1] + j * b[1] == new[1])) &amp; (i + j &gt;0): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; result = i + j print(result) &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;

不知道这个怎么处理的？

知道了 &nbsp;在tema &lt;= n &nbsp; temb &gt;= m

<pre class="prettyprint lang-py linenums">step = ((1,2),(2,1),(-1,-2),(-1,2),(1,-2),(-2,-1),(-2,1),(2,-1))
Isvisit = [[False]*(m+1) for _ in range(n+1)]
Isvisit[0][0] = True
visit = [(0,0)]
def f(n,m):
 minCount = 0
 while len(visit) != 0:
 l = len(visit)
 minCount += 1
 for i in xrange(l):
 a,b = visit.pop(0)
 for temp in step:
 tema = a + temp[0]
 temb = b + temp[1]
 if tema == n and temb == m:
 return minCount
 if tema &gt; -1 and tema &lt; n and temb &gt;-1 and temb &lt; m and not Isvisit[tema][temb]:
 Isvisit[tema][temb] == True
 visit.append((tema,temb))
 else:
 return -1
print f(n,m)
 
 </pre>

通不过，但是在编译器可以

<pre style="font-family: 宋体; font-size: 10.5pt;">def count_ri((m,n)):
 (x,y)=(0,0)
 count = 0
 while(m != x or n != y):
 #离目标很远
 if(((m-x&gt;=2) and (n-y&gt;=1)) or ((m-x&gt;=1) and (n-y&gt;=2))):
 if((m-x) &gt;= (n-y)):
 (x,y) = (x+2,y+1)
 count = count + 1
 if((m-x) &lt; (n-y)):
 (x,y) = (x+1,y+2)
 count = count + 1
 #离目标（1，1）
 elif(m-x==1 and n-y==1):
 (x,y) = (m,n)
 count = count + 4
 #离目标（1，0）或（0，1）
 elif((m-x==1 and n==y) or (n-y==1 and m==x)):
 (x,y) = (m,n)
 count = count + 3
 #离目标（2，0）或（0，2）
 elif((m-x==2 and n==y) or (n-y==2 and m==x)):
 (x,y) = (m,n)
 count = count + 2
 #离目标（远，0）或（0，远）
 elif((m-x&gt;2 and n==y) or (n-y&gt;2 and m==x)):
 if(m-x&gt;2 and n==y):
 (x,y) = (x+2,y-1)
 count = count + 1
 else:
 (x,y) = (x-1,y+2)
 count = count + 1
 #输出操作步数
 print count

count = 0
flag = 1
while(flag):
 if(m==1):
 if(n%4 ==2):
 count = n/2
 print count
 else:
 print -1
 break
 elif(n==1):
 if(m%4 ==2):
 count = m/2
 print count
 else:
 print -1
 break
 elif(n==2 and m==2):
 print -1
 break
 else:
 count_ri((m,n))
 break</pre>

pycharm测试没 问题，online提交，说变量有问题。请大神看看

from collections import deque def next_point(point, move): &nbsp;&nbsp;&nbsp; ''' &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 返回下一步坐标 &nbsp;&nbsp;&nbsp; ''' &nbsp;&nbsp;&nbsp; return (point[0]+move[0], point[1]+move[1], point[2]+1) &nbsp;&nbsp; &nbsp; def countWay(n, m, start=(0,0,0)): &nbsp;&nbsp;&nbsp; # direct存储特定方向对应的移动情况，（2，1）则表示 n方向+2, m方向+1， n方向为垂直方向，m方向为水平方向 &nbsp;&nbsp;&nbsp; direct = dict(zip(range(8), [(2,1), (1,2), (-1,2), (-2,1), (-2, -1), (-1,-2), (1,-2), (2, -1)])) &nbsp;&nbsp;&nbsp; # start[0]表示n方向坐标 start[1]表示y方向坐标 &nbsp;&nbsp;&nbsp; q = deque([start]) &nbsp;&nbsp;&nbsp; while len(q)&gt;0: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; point = q.popleft() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; for i in direct.keys(): &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; next_pos = next_point(point, direct[i]) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if 0&lt;=next_pos[0]&lt;=n and 0&lt;=next_pos[1]&lt;=m: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; q.append(next_pos) &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; if next_pos[0]==n and next_pos[1]==m: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; return next_pos[2] &nbsp;&nbsp;&nbsp; return -1 &nbsp;&nbsp; &nbsp; print(countWay(n,m))

出现超时，不知如何解决，有没有谁能提下修改意见？

修改后 &nbsp;避免反复步骤导致的死循环 ，但是 没有AC &nbsp;哪里错误？ def list_add(a,b): &nbsp; &nbsp; &nbsp; &nbsp; return [a[i]+b[i] for i in range(0,max(len(a),len(b)))] def comp_list(a,b): &nbsp; &nbsp; &nbsp; &nbsp; if a[0]&gt;b[0] or a[0]&lt;0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; &nbsp; &nbsp; if a[1]&gt;b[1] or a[1]&lt;0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; &nbsp; &nbsp; if a[0]==b[0]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if a[1] == b[1]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return False &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return True def BFS(org,ac,des): &nbsp; &nbsp; &nbsp; &nbsp; count = 0 &nbsp; &nbsp; &nbsp; &nbsp; re=[] &nbsp; &nbsp; &nbsp; &nbsp; stack=[] &nbsp; &nbsp; &nbsp; &nbsp; re.append(org) &nbsp; &nbsp; &nbsp; &nbsp; history=[] &nbsp; &nbsp; &nbsp; &nbsp; while len(re)&gt;0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tmp=re.pop(len(re) - 1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; count +=1 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; for i in ac: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; next_ac=list_add(tmp,i) &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if comp_list(next_ac,des): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; if history!=next_ac: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; stack.append(next_ac) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; elif next_ac[0]==des[0] and next_ac[1] == des[1]: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; return count &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; #re=re+stack &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; #stack = [] &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; history = tmp &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; while len(stack)&gt;0: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tmp2 = &nbsp;stack.pop(len(stack)-1) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; re.append(tmp2) &nbsp; &nbsp; &nbsp; &nbsp; return -1 ac=[[1,2],[2,1],[2,-1],[1,-2],[-1,-2],[-2,-1],[-2,1],[-1,2]] org=[0,0] des=[15,12] print BFS(org,ac,des)

def list_add(a,b): 	return [a[i]+b[i] for i in range(0,max(len(a),len(b)))] def comp_list(a,b): 	if a[0]&gt;b[0] or a[0]&lt;0: 		return False 	if a[1]&gt;b[1] or a[1]&lt;0: 		return False 	if a[0]==b[0]: 		if a[1] == b[1]: 			return False 	else: 		return True def BFS(org,ac,des): 	count = 1 	re=[] 	stack=[] 	re.append(org) 	while len(re)&gt;0: 		tmp=re.pop(len(re) - 1) 		for i in ac: 			next_ac=list_add(tmp,i) &nbsp; 			if comp_list(next_ac,des): 				stack.append(next_ac) 				count +=1 			elif next_ac[0]==des[0] and next_ac[1] == des[1]: 				return count 		while len(stack)&gt;0: 			tmp2 = &nbsp;stack.pop(len(stack)-1) 			re.append(tmp2) 	return -1 ac=[[1,2],[2,1],[2,-1],[1,-2],[-1,-2],[-2,-1],[-2,1],[-1,2]] org=[0,0] des=[n,m] print BFS(org,ac,des)

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