关于#python#的问题：如果两侧河岸上的狮子数量大于羚羊数量

Question

写一个程序来解决这样一个问题：3 只羚羊和3 只狮子准备乘船过河，河边有一艘能容纳2 只动物的小船。但是，如果两侧河岸上的狮子数量大于羚羊数量，羚羊就会被吃掉。
找到运送办法，使得所有动物都能安全渡河。
每个循环每个分支每个变量可以有注释

Answer 1

代码的输出结果含义是小船每次一来一回载的动物

from collections import Counter
from random import sample, randint
class Solution:
    """解决问题的方案类"""

    def __init__(self):
        """初始化属性"""
        self.left = ['羚羊'] * 3 + ['狮子'] * 3  # 后面用到了Counter，所以这里可以用字符串表示，不用0,1表示，更直观一点
        self.left_checkpoint = []  # 左边的存档，用于试错后恢复
        self.right = []
        self.right_checkpoint = []
        self.result = [[]]  # 结果，给个初始值是为了避免out of index的情况，取结果的时候切片即可
        self.result_checkpoint = []
        self.r_direction = True  # True为右，False为左

    def go(self):
        """渡河"""
        if self.r_direction:  # 向右渡河
            boat = sample(self.left, 2)
            for i in boat:
                self.left.remove(i)
                self.right.append(i)
        else:  # 向左渡河
            if len(self.right) > 1:  # 这里判断是为了避免sample取的时候越界(从1个里面取2个)
                boat = sample(self.right, randint(1, 2))
            else:
                boat = sample(self.right, 1)
            for i in boat:
                self.right.remove(i)
                self.left.append(i)
        return boat

    def judge(self):
        """判断"""
        if self.left and self.right:
            left_counter = Counter(self.left)
            right_counter = Counter(self.right)
            if (left_counter['羚羊'] and left_counter['羚羊'] < left_counter['狮子']) or \
                    (right_counter['羚羊'] and right_counter['羚羊'] < right_counter['狮子']):
                return False
        return True

    def checkpoint(self):
        """检查点"""
        self.left_checkpoint, self.right_checkpoint, self.result_checkpoint = \
            self.left.copy(), self.right.copy(), self.result.copy()

    def reset(self):
        """读档"""
        self.left, self.right, self.result = \
            self.left_checkpoint.copy(), self.right_checkpoint.copy(), self.result_checkpoint.copy()

    def get_result(self):
        """模拟渡河过程，获取结果"""
        while len(self.right) < 6:
            self.checkpoint()  # 存档
            boat = self.go()  # 渡河
            boat.sort()
            if self.judge() and boat != self.result[-1]:  # 这里判断是为了避免相同的人来回的情况，以求尽可能少的解
                self.r_direction = not self.r_direction  # 调转船头
                self.result.append(boat)
            else:
                self.reset()  # 读档
        return self.result[1:]


def main():
    """主函数"""

    repeat = 10000  # 重复执行次数
    result_set = set()  # 解的集合
    solution = Solution()

    for _ in range(repeat):
        result = solution.get_result()
        result_set.add(str(result))
        solution.__init__()

    print(f'经{repeat}次执行，共得到{len(result_set)}种不同的结果，结果如下：', end='\n\n')
    for result in result_set:

        print(result)
if __name__ == '__main__':
    main()