帮我编写一个类似吃鸡游戏的源代码继续优化

下面是一个简单的吃鸡游戏源代码，你可以根据自己的需求进行进一步的优化：

import random

class Player:
    def __init__(self, name):
        self.name = name
        self.health = 100
        self.attack_power = 10
        self.defense_power = 5
        self.money = 0

    def attack(self, enemy):
        damage = random.randint(1, self.attack_power)
        enemy.defend(damage)

    def defend(self, damage):
        actual_damage = damage - self.defense_power
        self.health -= actual_damage
        if self.health <= 0:
            print(f"{self.name}被击败了！")
            exit()

    def collect_money(self, amount):
        self.money += amount

class Enemy:
    def __init__(self, name, health, attack_power, defense_power):
        self.name = name
        self.health = health
        self.attack_power = attack_power
        self.defense_power = defense_power

    def attack(self, player):
        damage = random.randint(1, self.attack_power)
        player.defend(damage)

    def defend(self, damage):
        actual_damage = damage - self.defense_power
        self.health -= actual_damage
        if self.health <= 0:
            print(f"{self.name}被击败了！")
            player.collect_money(random.randint(1, 10))

player_name = input("请输入玩家名称：")
player = Player(player_name)

enemies = [Enemy("敌人1", 50, 5, 2), Enemy("敌人2", 60, 6, 3), Enemy("敌人3", 70, 7, 4)]

while True:
    enemy = random.choice(enemies)
    print(f"你遇到了{enemy.name}！")
    action = input("请选择你的行动（1. 攻击；2. 逃跑）：")
    if action == "1":
        player.attack(enemy)
        enemy.attack(player)
    elif action == "2":
        print("你逃跑了！")
        break
    else:
        print("无效的选择，请重新选择。")

    print(f"你的健康值：{player.health}；你的金币数：{player.money}")

这段代码实现了一个简单的吃鸡游戏，玩家可以选择攻击敌人或逃跑。玩家和敌人都有健康值、攻击力和防御力。当玩家的健康值降到0或以下时，玩家被击败。当敌人的健康值降到0或以下时，敌人被击败，玩家会获得随机数量的金币。游戏会不断循环，直到玩家选择逃跑为止。你可以根据自己的需求进一步优化代码，例如增加更多的敌人、道具和技能等