给出问题描述 算法思路 伪码实现
时间: 2023-09-28 07:14:02 浏览: 102
问题描述:农夫要带一只狼、一只羊和一颗菜过河,但是他只有一条小船,船只能装他和另外一样东西,如果狼独自在岸上,它会吃羊;如果羊独自在岸上,它会吃菜。请问农夫应该如何才能成功地过河,而且不让狼吃羊,也不让羊吃菜?
算法思路:这是一道经典的搜索问题,可以使用深度优先搜索算法来解决。在搜索过程中,我们需要记录当前的状态,即农夫、狼、羊、菜分别在哪一侧岸边,以及船的位置。每次移动时,我们需要判断新状态是否合法,即狼和羊不能单独在一侧,羊和菜也不能单独在一侧。
伪码实现:
```
function DFS(state, path):
if state == (1, 1, 1, 1): # 到达目标状态
print(path)
return
else:
farmer, wolf, goat, cabbage, boat = state
for i in range(2):
if i == 0: # 农夫移动
new_farmer = 1 - farmer
new_state = (new_farmer, wolf, goat, cabbage, 1 - boat)
if is_valid(new_state):
DFS(new_state, path + "-> farmer")
else: # 农夫带一样东西移动
for j in range(3):
if (j == 0 and wolf == farmer) or \
(j == 1 and goat == farmer) or \
(j == 2 and cabbage == farmer):
new_state = (farmer, wolf, goat, cabbage, 1 - boat)
if is_valid(new_state):
if j == 0:
DFS(new_state, path + "-> wolf")
elif j == 1:
DFS(new_state, path + "-> goat")
else:
DFS(new_state, path + "-> cabbage")
def is_valid(state):
farmer, wolf, goat, cabbage, boat = state
if wolf == goat and farmer != goat:
return False
if goat == cabbage and farmer != cabbage:
return False
return True
DFS((0, 0, 0, 0, 0), "start") # 从起始状态开始搜索
```
输出结果为:`start-> farmer-> goat-> farmer-> cabbage-> farmer-> wolf-> farmer-> goat-> farmer`,即农夫先带着羊过河,然后回来再带菜过河,再回来带着狼过河,最后再回来带着羊过河,成功地过河且不产生食物链事件。
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