菜鸡狼过河python
时间: 2023-11-15 21:01:51 浏览: 56
这是一个经典的问题,也称为“狼、羊、菜和农夫过河问题”。在这个问题中,农夫需要带着一只狼、一只羊和一些菜过河,但是船只只能容纳农夫和另外一只物品。如果农夫不在场,狼会吃羊,羊会吃菜。因此,农夫需要找到一种方法,使得所有物品都能安全地过河。
以下是一个简单的Python实现,其中使用了深度优先搜索算法来解决问题:
```
# 定义初始状态和目标状态
start_state = ['left', 'left', 'left', 'left']
goal_state = ['right', 'right', 'right', 'right']
# 定义有效状态
def is_valid(state):
if state[0] != state[1] and state[1] != state[2]:
return True
return False
# 定义搜索函数
def dfs(state, path):
if state == goal_state:
return path
for i in range(8):
new_state = state.copy()
if i == 0:
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 1:
if new_state[3] == new_state[0]:
new_state[1] = 'right' if new_state[1] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 2:
if new_state[3] == new_state[1]:
new_state[0] = 'right' if new_state[0] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 3:
if new_state[3] == new_state[2]:
new_state[2] = 'right' if new_state[2] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 4:
if new_state[3] == 'right':
new_state[3] = 'left'
elif i == 5:
if new_state[3] == new_state[0]:
new_state[1] = 'right' if new_state[1] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 6:
if new_state[3] == new_state[1]:
new_state[0] = 'right' if new_state[0] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
elif i == 7:
if new_state[3] == new_state[2]:
new_state[2] = 'right' if new_state[2] == 'left' else 'left'
new_state[3] = 'right' if new_state[3] == 'left' else 'left'
if is_valid(new_state) and new_state not in path:
new_path = dfs(new_state, path + [new_state])
if new_path:
return new_path
return None
# 执行搜索并输出结果
result = dfs(start_state, [start_state])
if result:
for state in result:
print(state)
else:
print('No solution found.')
```