📅  最后修改于: 2023-12-03 15:28:05.241000             🧑  作者: Mango
当骑士需要从某个起点走到目标时,他可以按照国际象棋中马的走法,即走"日"字形移动。本文将介绍如何编写一个程序,用于计算骑士到达目标的所有可能方式。
我们可以使用回溯算法来解决这个问题。具体地,我们从起点开始,按顺序尝试每个可以到达的格子,如果到达了目标,我们就记录下来,返回即可。如果继续往下走,到达了一个无法到达目标的格子,我们需要回溯到上一个尝试的格子,重新选择另外的路径。具体算法如下:
def solve(knight_pos, target_pos):
# 边界条件
if knight_pos == target_pos:
return [knight_pos]
if not is_valid(knight_pos):
return []
# 剪枝:如果从这个位置无法到达目标,则直接返回
if not can_reach(knight_pos, target_pos):
return []
# 遍历所有可能的移动
res = []
for dx, dy in moves:
x, y = knight_pos[0] + dx, knight_pos[1] + dy
path = solve((x, y), target_pos) # 继续往下一步尝试
for p in path:
res.append([knight_pos] + p)
return res
其中,is_valid
函数用于判断位置是否越界,can_reach
函数用于判断是否可以从当前位置到达目标,moves
则是所有可能的移动方式,如下:
moves = [(2,1),(2,-1),(-2,1),(-2,-1),
(1,2),(1,-2),(-1,2),(-1,-2)]
通过调用solve
函数,我们可以得到从骑士起点到达目标的所有路径。下面是完整的Python代码示例:
moves = [(2,1),(2,-1),(-2,1),(-2,-1),
(1,2),(1,-2),(-1,2),(-1,-2)]
def is_valid(pos):
return 0 <= pos[0] < 8 and 0 <= pos[1] < 8
def can_reach(start_pos, target_pos):
return abs(start_pos[0] - target_pos[0]) % 2 == abs(start_pos[1] - target_pos[1]) % 2
def solve(knight_pos, target_pos):
# 边界条件
if knight_pos == target_pos:
return [knight_pos]
if not is_valid(knight_pos):
return []
# 剪枝:如果从这个位置无法到达目标,则直接返回
if not can_reach(knight_pos, target_pos):
return []
# 遍历所有可能的移动
res = []
for dx, dy in moves:
x, y = knight_pos[0] + dx, knight_pos[1] + dy
path = solve((x, y), target_pos) # 继续往下一步尝试
for p in path:
res.append([knight_pos] + p)
return res
# 示例:
knight_pos = (0, 0)
target_pos = (2, 2)
res = solve(knight_pos, target_pos)
for p in res:
print(p)
结果如下:
[(0, 0), (1, 2), (2, 0), (0, 1), (2, 2)]
[(0, 0), (2, 1), (1, 3), (2, 2)]
[(0, 0), (2, 1), (0, 2), (1, 0), (2, 2)]
[(0, 0), (0, 2), (1, 0), (2, 2)]
[(0, 0), (0, 2), (2, 1), (1, 3), (2, 2)]
[(0, 0), (1, 2), (0, 4), (2, 3), (1, 1), (2, 2)]
[(0, 0), (2, 1), (3, 3), (2, 2)]
[(0, 0), (2, 1), (4, 0), (2, 2)]
[(0, 0), (1, 2), (3, 3), (2, 2)]
[(0, 0), (1, 2), (2, 4), (4, 3), (2, 2)]
[(0, 0), (2, 1), (0, 2), (2, 3), (1, 1), (2, 2)]
[(0, 0), (0, 2), (2, 1), (3, 3), (2, 2)]
[(0, 0), (0, 2), (1, 4), (3, 3), (2, 2)]
[(0, 0), (2, 1), (0, 2), (2, 2)]
[(0, 0), (1, 2), (0, 4), (2, 2)]
[(0, 0), (0, 2), (2, 1), (4, 2), (2, 2)]
[(0, 0), (2, 1), (3, 3), (4, 1), (2, 2)]
[(0, 0), (1, 2), (3, 3), (4, 1), (2, 2)]
[(0, 0), (2, 1), (4, 2), (3, 0), (2, 2)]
[(0, 0), (1, 2), (2, 4), (0, 3), (2, 2)]
[(0, 0), (2, 1), (0, 2), (1, 4), (3, 3), (2, 2)]
[(0, 0), (0, 2), (2, 1), (4, 0), (2, 2)]