Python 初学笔记：递归解决0-1背包问题

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最近看了Magnus Lie Hetland的《Python基础教程（第2版）》158页利用生成器和递归解决八皇后问题的内容，于是想用这种方法解决0/1背包问题，所以仿照基础教程写了一段代码，基本上能够解决问题，但是明显还有不足之处，先记录下，以后慢慢修改。。。

def conflict(m, w, state, pos):    total = 0    for i in state:        total += w[i]    if total <= m and total + w[pos] > m:        return True    else:        return Falsedef knapsacks(m, w, state=()):    n = len(w)    for i in range(n):        if i not in state:            if not conflict(m, w, state, i):                for result in knapsacks(m, w, state + (i,)):                    yield (i,) + result            else:                yield ()def max_knapsack(m = 10, w = (2,2,6,5,4), p = (6,3,5,4,6)):    knss = list(knapsacks(m, w))    ps = []    for kns in knss:        total = 0        for k in kns:            total += p[k]        ps.append(total)    return max(ps)def main():    m = raw_input("Input weight constraint of knapsacks: ")    n = raw_input("Input the number of goods: ")    w = raw_input("Input each weight of the %s goods (e.g., 2,3,5,...): " % n)    p = raw_input("Input each value of the %s goods (e.g., 2,3,5,...): " % n)    m = float(m)    n = int(n)    ws = w.split(',')    ps = p.split(',')    w = []    p = []    for i in range(n):        w.append(float(ws[i]))        p.append(float(ps[i]))    print "Max value: ", max_knapsack(m, w, p)main()

从结果上明显可看出knapsacks和conflict方法还有一些不足之处。

knapsacks方法返回的可能所有满足总重量小于m的组合：

由于递归方法写得不好，导致出现多个重复的元组。

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