动态编程

很多问题可以分解为简单的迭代，迭代过程中回重复计算一些已经计算过的值，可以把原来计算过的值存放到一个字典中，这样可以大大提高迭代的效率

这里以斐波那契数列和背包问题来展现动态编程的应用

# MIT600《计算机科学及编程导论》（2008年秋季）样码
# 第十三讲
# 翻译制作：ocourse.org
# 课程讨论版：http://ocourse.org/bbs/forum.php?mod=forumdisplay&fid=29
# by yoeo24

def fib(n):
    global numCalls
    numCalls +=1
    #print 'fib调用', n
    if n<=1:
        return 1
    else:
        return fib(n-1)+fib(n-2)
numCalls = 0
n = 6
res = fib(n)
print 'fib_', n, '=', res, '调用次数为', numCalls

def fastFib(n, memo):
    global numCalls
    numCalls +=1
    #print 'fib1调用', n
    if not n in memo:
        memo[n]=fastFib(n-1, memo) + fastFib(n-2, memo)
    return memo[n]

def fib1(n):
    memo = {0:1, 1:1}
    return fastFib(n, memo)

numCalls = 0
n = 6
res = fib1(n)
print 'fib_', n, '=', res, '调用次数为', numCalls
    
def maxVal(w,v,i,aW):
    #print 'maxVal调用:', i, aW
    global numCalls
    numCalls +=1
    if i == 0:
        if w[i] <= aW: return v[i]
        else: return 0
    without_i = maxVal(w,v,i-1,aW)
    if w[i] > aW:
        return without_i
    else:
        with_i = v[i] + maxVal(w,v,i-1,aW-w[i])
    return max(with_i, without_i)

weights = [1,5,3,4]
vals = [15,10,9,5]
numCalls = 0
res = maxVal(weights, vals, len(vals)-1, 8)
print '最大价值是',res, '调用次数', numCalls

weights = [1,1,5,5,3,3,4,4]
vals = [15,15,10,10,9,9,5,5]
numCalls = 0
res = maxVal(weights, vals, len(vals)-1, 8)
print '最大价值是',res, '调用次数', numCalls

maxWeight = 40
w = [5,5,1,8,2,9,7,5,2,8,1,2,7,3,5,7,8,5,5,8,2,3,8,4,9]
v = [5,5,3,5,6,7,2,1,7,1,8,3,6,8,8,6,5,6,8,4,3,5,2,3,4] #看不全
numCalls = 0
res = maxVal(w, v, len(v)-1, maxWeight)
print '最大价值是',res, '调用次数', numCalls

def fastMaxVal(w,v,i,aW,m):
    global numCalls
    numCalls +=1
    try: return m[(i,aW)]
    except KeyError:
        if i == 0:
            if w[i] <= aW:
                m[(i,aW)]=v[i]
                return v[i]
            else:
                m[(i,aW)]=0
                return 0
        without_i = fastMaxVal(w,v,i-1,aW,m)
        if w[i] > aW:
            m[(i,aW)] = without_i
            return without_i
        else:
            with_i = v[i]+fastMaxVal(w,v,i-1,aW-w[i],m)
        res = max(with_i,without_i)
        m[(i,aW)]=res
        return res

def maxVal0(w,v,i,aW):
    m={}
    return fastMaxVal(w,v,i,aW,m)

weights = [1,1,5,5,3,3,4,4]
vals = [15,15,10,10,9,9,5,5]
numCalls = 0
res = maxVal0(weights, vals, len(vals)-1, 8)
print '最大价值是',res, '调用次数', numCalls

maxWeight = 40
w = [5,5,1,8,2,9,7,5,2,8,1,2,7,3,5,7,8,5,5,8,2,3,8,4,9]
v = [5,5,3,5,6,7,2,1,7,1,8,3,6,8,8,6,5,6,8,4,3,5,2,3,4] #看不全
numCalls = 0
res = maxVal0(w, v, len(v)-1, maxWeight)
print '最大价值是',res, '调用次数', numCalls

背包问题：

w 为重量

v 为价值

maxWeight为背包容量

以深度优先，左边优先遍历树，并通过回溯来解决问题