August 2017
Intermediate to advanced
360 pages
8h 35m
Chinese
Content preview from 算法技术手册（原书第2 版）
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306
｜
第
11
章
remaining = j - weights[i-1]
m[i][j] = max(m[i-1][j], m[i-1][remaining] + values[i-1])
else
m[i][j] = m[i-1][j]
return m[n][W]
❹
end
Knapsack unbounded(weights, values, W)
n = number of items
m = empty (W+1) vector
❺
for j=1 to W+1 do
best = m[j-1]
for i=0 to n-1 do
remaining = j - weights[i]
if remaining >= 0 and m[remaining] + values[i] > best then
best = m[remaining] + values[i]
m[j] = best
return m[W]
❶
m
[
i
][
j
]
记录的是考虑
i
个物品而且不超过质量
w
时能够获得的最大价值。
❷
能否通过将之前的将第
i
－
1
号物品加入背包呢？
❸
第
i
－
1
号物品超过了质量限制，所以不能改进解决方案。
❹
返回计算出的最佳结果。
❺
对于无限背包问题来说，
m
[
j
]
记录的是不超过质量
j
的最大物品价值。
11.2.1 输入 / 输出
给定一系列物品（每件物品都有一个整数值的质量和价值） ...
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