问题描述
- 大盗潜入博物馆,面前有5件宝物,分别有重量和价值,大盗的背包仅能附中20公斤,请问如何选择宝物,是的总价值最高?
- 宝物重量、价值对应关系表格如下:
item |
weight |
value |
1 |
2 |
3 |
2 |
3 |
4 |
3 |
4 |
8 |
4 |
5 |
8 |
5 |
9 |
10 |
解题思路
- 将 m ( i , W ) m(i, W) m(i,W)标记为:
- 前 i ( 1 < = i < = 5 ) i(1<=i<=5) i(1<=i<=5)个宝物中,组合不超过 W ( 1 < = W < = 20 ) W(1<=W<=20) W(1<=W<=20)重量,得到的最大价值
- m ( i , W ) m(i,W) m(i,W)应该是 m ( i − 1 , W ) m(i-1,W) m(i−1,W)和 m ( i − 1 , W − W i ) + V i m(i-1, W-W_{i})+V_{i} m(i−1,W−Wi)+Vi,两者的最大值
- 从 m ( 1 , 1 ) m(1,1) m(1,1)开始计算到 m ( 5 , 5 ) m(5,5) m(5,5)
代码实现
def maxValue(max_w):
"""
:max_w: 最大携带重量
"""
tr = [None, {
'w':2, 'v':3}, {
'w':3, 'v':4},
{
'w':4, 'v':8}, {
'w':5, 'v':8},
{
'w':9, 'v':10}]
m = {
(i,w): 0 for i in range(len(tr))
for w in range(max_w + 1)}
for i in range(1, len(tr)):
for w in range(1, max_w + 1):
if tr[i]['w'] > w:
m[(i,w)] = m[(i-1, w)]
else:
m[(i,w)] = max(m[(i-1, w)], m[(i-1,w-tr[i]['w'])] + tr[i]['v'])
return m[(len(tr)-1, max_w)]
maxValue(20)
递归的思想-代码实现
递归算法“三定律”
- 递归算法必须具备基本结束条件
- 递归算法必须要减小规模,改变状态,向基本结束条件演进
- 递归算法必须调用自身
tr = {
(2,3), (3,4), (4,8),(5,8), (9,10)}
max_w = 20
m = {
}
def thief(tr, w):
if tr == set() or w == 0:
m[tuple(tr), w] = 0
return 0
elif (tuple(tr), w) in m:
return m[tuple(tr), w]
else:
vmax = 0
for t in tr:
if t[0] <= w:
v = thief(tr-{
t}, w-t[0]) + t[1]
vmax = max(vmax, v)
m[tuple(tr), w] = vmax
return vmax
print(thief(tr, max_w))