游戏水桶倒水问题Python语言的解决方法，倒水python,Python编程语言关于

游戏水桶倒水问题Python语言的解决方法，倒水python,Python编程语言关于

Python编程语言关于游戏水桶倒水问题的解决方法是怎么样的？问题是：有三个水桶，分别是10升、7升和3升。10升的水桶是满的，要求想办法把另外两个桶变成5升的，并且不能浪费水哦！需要用Python来完成这个游戏，大家有什么好的思路吗？

其他Python相关游戏源码文章推荐：Python实现自动扫雷游戏的方法源码

下边先来看看我的水桶倒水问题Python语言的解决方法：

#! /usr/bin/env python# -*- coding: utf-8 -*-LIMITS = (10, 7, 3)INIT = (10, 0, 0)WIN = (5, 5, 0)from copy import copyclass State:    def __init__(self, init_value):        self.parent = -1        self.value = init_value        self.move = (0, 0, 0) #(from, to, amount)    #spawn by put water around    def spawn(self):        for i in range(len(LIMITS)): #for each cup            if not self.value[i]: continue #if empty cup then next            for k in range(len(LIMITS)): #otherwise try put water to other cups                if i == k: continue                val_i = self.value[i]                val_k = self.value[k]                #1. move all in i to k                #2. move i to make k full                #so the minimum of the 2 actions is the target                min_ik = min(val_i, LIMITS[k] - val_k)                new_value = list(copy(self.value))                new_value[i] -= min_ik                new_value[k] += min_ik                new_value = tuple(new_value)                if new_value in g_state_map: #already exists?                    continue                new_state = State(new_value)                new_state.parent = g_index                new_state.move = (i, k, min_ik)                g_states.append(new_state)                g_state_map[new_value] = True                if new_value == WIN:                    return True        return False#www.iplaypy.comdef print_solution(state):    states = []    while state.parent != -1:        states.append(state)        state = g_states[state.parent]    states.reverse()    print "At least:", len(states), "steps:"    print "(Cup0, Cup1, Cup2) (from, to, amount)"    print INIT    for state in states:        print state.value, state.move#solve the problem here!g_states = [State(INIT)] #initially we have (10, 0, 0)g_state_map = {} #sate:true/false, avoid duplicate statesg_index = 0while g_index < len(g_states):    state = g_states[g_index]    if state.spawn():        print_solution(g_states[-1])        break    g_index += 1result = """At least: 9 steps:(Cup0, Cup1, Cup2) (from, to, amount)(10, 0, 0)(3, 7, 0) (0, 1, 7)(3, 4, 3) (1, 2, 3)(6, 4, 0) (2, 0, 3)(6, 1, 3) (1, 2, 3)(9, 1, 0) (2, 0, 3)(9, 0, 1) (1, 2, 1)(2, 7, 1) (0, 1, 7)(2, 5, 3) (1, 2, 2)(5, 5, 0) (2, 0, 3)"""

python代码中的python reverse方法是列表反转排序，具体使用方法可点击链接查看。

编橙之家文章，