编辑代码

# coding:utf-8
#JSRUN引擎2.0，支持多达30种语言在线运行，全仿真在线交互输入输出。 
from collections import defaultdict

def dfs():
    # 定义三个瓶子的容量：A=10两, B=7两, C=3两
    capacities = (10, 7, 3)
    # 初始状态：A 满，B、C 空
    initial = (10, 0, 0)
    # 记录已访问的状态，防止重复搜索
    visited = set([initial])
    # 用字典记录每个状态的前驱状态和进行的倒油操作
    # 格式：state: (prev_state, (from, to, pour_amount))
    parent = {initial: None}
    move_taken = {initial: None}

    def dfs_helper(state):
        # 判断是否达到目标状态：A 瓶中有5两
        if state[0] == 5:
            return state
        # 遍历所有可能的倒油操作（从 i 倒到 j）
        for i in range(3):
            for j in range(3):
                if i == j:
                    continue
                # 如果 i 瓶中没有油或者 j 瓶已满则跳过
                if state[i] == 0 or state[j] == capacities[j]:
                    continue
                # 计算本次能倒出的油量：取 i 瓶剩余和 j 瓶可装容量中的较小者
                amount = min(state[i], capacities[j] - state[j])
                # 生成新的状态
                new_state = list(state)
                new_state[i] -= amount
                new_state[j] += amount
                new_state = tuple(new_state)

                if new_state not in visited:
                    visited.add(new_state)
                    parent[new_state] = state
                    move_taken[new_state] = (i, j, amount)
                    result = dfs_helper(new_state)
                    if result:
                        return result
        return None

    final_state = dfs_helper(initial)
    if not final_state:
        return None
    # 重构路径（倒序回溯到初始状态）
    path = []
    cur = final_state
    while cur is not None:
        path.append((cur, move_taken[cur]))
        cur = parent[cur]
    path.reverse()
    return path

def print_solution(solution):
    if not solution:
        print("未找到解法")
        return
    print("初始状态为 (A, B, C)：(10, 0, 0)")
    for idx, (state, move) in enumerate(solution):
        # 初始状态没有倒油操作
        if move is None:
            continue
        from_idx, to_idx, amount = move
        # 对应的瓶子名称
        names = ['A(10两)', 'B(7两)', 'C(3两)']
        print(f"第{idx}步：从 {names[from_idx]} 倒 {amount} 两 到 {names[to_idx]}，得到状态 {state}")
    print("目标状态：A 瓶中正好有 5 两油，实现均分！")

if __name__ == '__main__':
    solution = dfs()
    print_solution(solution)