题目

某旅行者外出, 需要将5件物品装入包中. 包的总容量是12kg, 物品重量及价值如表. 问如何装这些物品, 才能使得总价值最大? (写出递归式, 列表计算, 算法).

item	A	B	C	D	E
weight	11	5	4	3	1
value	8	4	3	2	0.5

解答

(1) Optimal Substructure

对某个物品来说, 无论取或者不取, 余下容量的装填都应该是最优的.

(2) Recurrence Equation

在决策第i个物品的决策时, i~j物品的价值总和S有:
S(i, j, Cap) = max{ S(i-1, j, Cap-w[i])+v[i], S(i-1, j, Cap)},
其中, Cap表示剩余容量Capacity;

注意到j实际上是固定的值, 即j=n, 所以可以简化成:

S(i, Cap) = max{ S(i-1, Cap-w[i])+v[i], S(i-1, Cap) }

考虑边界情况,

a. 当w[i] > Cap时, 物品已经超出容量, 因此:

S(i, Cap) = S(i-1, Cap)

b. 当i>j, 或者Cap = 0时,

S(i, Cap) = 0

(3) Bottom-up Tabular Calculation

表格计算

tabular calculation.png

(4) Algorithm

knap(n, w[1~n], W)
#init
let S be a new table, rec be an array
for cap = 0~W:  #i > j
    S[6][cap] = 0
for i = 1~n+1:  #cap = 0
    S[i][0] = 0
#calculation
for i = n~1:
    for cap = 0~W:
        rec[i] = 0  #item[i] out by default
        if w[i] <= cap: #weight OK
            if S[i-1][cap-w[i]]+v[i] > S[i-1][cap]:
                S[i][cap] = S[i-1][cap-w[i]]+v[i]
                rec[i] = 1  #item[i] in
            else:
                S[i][cap] =  S[i-1][cap]
        else:  #too weighty
            S[i][cap] = S[i-1][cap]
return S, rec