农夫过河问题python实现

农夫过河问题是一个经典的动态规划（Dynamic Programming, DP）问题，通常涉及将动物（如牛、羊和鸡）从一边的岸运送到另一边，同时考虑每种动物单独过河需要船员陪同，而马可以同时承载其他两个动物过河。这里我们给出一个简单的Python实现：

def ferry-crossing(farmyard, limit=1):
    dp = [[0] * (farmyard + 1) for _ in range(farmyard + 1)]
    
    # 初始化边界条件
    dp[0][0] = dp[0][1] = dp[1][0] = 1

    # 动态规划遍历
    for i in range(2, farmyard+1):
        for j in range(i+1):
            if j - i >= 0 and i - (j - 2) >= 0:  # 当前动物无法单独过河，只能由马带
                dp[i][j] = dp[j - i][j - i] + dp[i - (j - 2)][j]
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])

    return dp[farmyard][farmyard]

# 测试
print(ferry_crossing(4))  # 输出结果取决于农场里有多少种动物以及它们的数量

相关问题

农夫过河问题python

好的，农夫过河问题是一个经典的逻辑问题，题目描述如下：

一个农夫要带着一只狼、一只羊和一颗白菜过河，但是小船只能装下农夫和另外一件物品。如果农夫不在场，狼就会吃羊，羊不在场，白菜就会被吃。请问农夫如何才能安全地将这三样东西全部运到河的对岸？

以下是一个简单的 Python 实现：

class State:
    def __init__(self, farmer, wolf, goat, cabbage, boat):
        self.farmer = farmer  # 0:在左岸，1:在右岸
        self.wolf = wolf
        self.goat = goat
        self.cabbage = cabbage
        self.boat = boat  # 0:在左岸，1:在右岸

    def is_valid(self):
        if self.farmer not in [0, 1] or self.wolf not in [0, 1] or self.goat not in [0, 1] or self.cabbage not in [0, 1] or self.boat not in [0, 1]:
            return False
        if self.wolf == self.goat and self.farmer != self.wolf:
            return False
        if self.goat == self.cabbage and self.farmer != self.goat:
            return False
        return True

    def is_goal(self):
        return self.farmer == 1 and self.wolf == 1 and self.goat == 1 and self.cabbage == 1

    def __eq__(self, other):
        return self.farmer == other.farmer and self.wolf == other.wolf and self.goat == other.goat and self.cabbage == other.cabbage and self.boat == other.boat

    def __hash__(self):
        return hash((self.farmer, self.wolf, self.goat, self.cabbage, self.boat))


class Node:
    def __init__(self, state, parent=None, action=None):
        self.state = state
        self.parent = parent
        self.action = action

    def expand(self):
        children = []
        for a in ["farmer", "wolf", "goat", "cabbage"]:
            new_state = State(self.state.farmer, self.state.wolf, self.state.goat, self.state.cabbage, self.state.boat)
            setattr(new_state, a, 1 - getattr(self.state, a))  # 改变属性值
            new_state.boat = 1 - self.state.boat  # 船也过河了
            child = Node(new_state, self, a)
            if new_state.is_valid():
                children.append(child)
        return children


def bfs(initial_state):
    initial_node = Node(initial_state)
    if initial_state.is_goal():
        return initial_node

    frontier = [initial_node]
    explored = set()

    while frontier:
        node = frontier.pop(0)
        explored.add(node.state)
        for child in node.expand():
            if child.state.is_goal():
                return child
            if child.state not in explored and child not in frontier:
                frontier.append(child)

    return None


initial_state = State(0, 0, 0, 0, 0)
solution = bfs(initial_state)

if not solution:
    print("No solution found!")
else:
    path = []
    node = solution
    while node:
        path.append(node)
        node = node.parent
    path.reverse()
    for node in path:
        print(node.state.farmer, node.state.wolf, node.state.goat, node.state.cabbage, node.state.boat, node.action)

这里使用了 BFS 算法来搜索解决方案，首先定义了 `State` 类来表示状态，包含农夫、狼、羊、白菜和船的位置，以及一些辅助方法。然后定义了 `Node` 类来表示搜索树中的节点，包含一个状态和一个指向父节点的指针。最后定义了 `bfs` 函数来执行 BFS 算法搜索解决方案，返回最终状态的节点。

农夫过河问题Python代码

农夫过河问题是一个经典的智力游戏，题目描述如下：
农夫要带着一只狼、一只羊和一棵白菜过河。但是，河边只有一条小船，而且这条小船最多只能载两个物品（包括农夫），而且狼会吃羊，羊会吃白菜。请问，农夫应该如何安排，才能安全地把这三个物品都运到对岸？

下面是一个可能的Python代码实现：

def farmer_cross_river(start, end):
    # 定义初始状态和目标状态
    initial_state = {'farmer': start, 'wolf': start, 'sheep': start, 'cabbage': start}
    goal_state = {'farmer': end, 'wolf': end, 'sheep': end, 'cabbage': end}
    # 定义安全状态
    safe_states = [
        {'farmer': start, 'wolf': start, 'sheep': start, 'cabbage': end},
        {'farmer': start, 'wolf': start, 'sheep': end, 'cabbage': start},
        {'farmer': end, 'wolf': end, 'sheep': start, 'cabbage': start},
        {'farmer': end, 'wolf': end, 'sheep': end, 'cabbage': end},
        {'farmer': end, 'wolf': start, 'sheep': end, 'cabbage': end}
    ]
    # 定义当前状态和已经走过的状态
    current_state = initial_state.copy()
    walked_states = [current_state]
    while current_state != goal_state:
        # 判断当前状态可以到达哪些状态
        next_states = []
        for item1 in current_state:
            if current_state[item1] == current_state['farmer']:
                for item2 in current_state:
                    if current_state[item2] == current_state[item1]:
                        next_state = current_state.copy()
                        next_state['farmer'] = end if current_state['farmer'] == start else start
                        next_state[item1] = end if current_state['farmer'] == start else start
                        next_state[item2] = end if current_state['farmer'] == start else start
                        if next_state in safe_states and next_state not in walked_states:
                            next_states.append(next_state)
        # 随机选择下一个状态
        if len(next_states) > 0:
            current_state = random.choice(next_states)
            walked_states.append(current_state)
        else:
            # 如果没有下一个状态可以到达，则回溯到上一个状态
            walked_states.pop()
            current_state = walked_states[-1]
    return walked_states

使用这个函数可以得到从起点到终点的所有可能状态的列表。