商人过河问题python代码

下面是一个用Python编写的商人过河问题的解决方案：

from itertools import combinations

def valid_state(state):
    # 检查状态是否合法
    if state[0] < state[1] and state[0] != 0:
        return False
    if state[2] < state[3] and state[2] != 0:
        return False
    return True

def solve():
    # 初始状态：(商人数量, 马数量, 目的地)
    initial_state = (3, 3, 1)
    # 目标状态：(0, 0, 0)
    target_state = (0, 0, 0)

    # 使用广度优先搜索
    queue = [(initial_state, [])]
    visited = set([initial_state])

    while queue:
        state, actions = queue.pop(0)

        if state == target_state:
            return actions

        m, c, b = state

        # 根据当前状态生成可能的下一步动作
        possible_actions = []
        if b == 1:
            possible_actions = [(x, y, 0) for x in range(m+1) for y in range(c+1) if 1 <= x + y <= 2]
        else:
            possible_actions = [(x, y, 1) for x in range(4-m, 4) for y in range(4-c, 4) if 1 <= x + y <= 2]

        # 遍历可能的下一步动作
        for action in possible_actions:
            if not valid_state(action):
                continue

            new_state = (m - action[0], c - action[1], action[2])

            if new_state not in visited:
                queue.append((new_state, actions + [action]))
                visited.add(new_state)

    return None

if __name__ == "__main__":
    actions = solve()
    if actions:
        for action in actions:
            if action[2] == 1:
                print(f"商人和马过河，左岸: {3-action[0]}个商人, {3-action[1]}匹马；右岸: {action[0]}个商人, {action[1]}匹马")
            else:
                print(f"商人和马返回，左岸: {3-action[0]}个商人, {3-action[1]}匹马；右岸: {action[0]}个商人, {action[1]}匹马")
    else:
        print("无解")

这段代码使用了广度优先搜索算法来解决商人过河问题。它首先定义了一个 `valid_state` 函数来检查状态是否合法。然后，使用一个队列 `queue` 来存储当前状态和到达该状态的动作序列。通过不断从队列中弹出状态并生成可能的下一步动作，直到找到目标状态或队列为空为止。

注意：这个代码只是一个简单的实现，并没有进行优化处理，可能在处理大规模问题时效率较低。