如何减少0〜1背包的时间复杂度

Question

As for 0~1 knapsack problem,

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

c[i] means the cost of ith goods, w[i] means the value of ith goods.

And I read one doc,which said the time complexity can be optimization,especially when V is larger.as below

 i=1...N

 v=V...0 

can be changed to

 i=1...n        

bound=max{V-sum{w[i..n]},c[i]}        

 v=V...bound

what does it mean?How can V(the maximum of bag) minus sum of w[i](the value of goods)?

Really confuse,or something wrong on this doc?

Answer 1

You didn't say whose complexity you are optimizing. Are you using dynamic programming? If so, this could mean that you don't need to calculate f[i][v] for small values of v, because you won't need those values to find the optimum.

Even if you put all goods from i + 1 to n in the knapsack, you still have a capacity of V - sum{c[i+1..n]} left, so you don't need to solve the subproblem i (restricted to goods 1..i ) with a capacity smaller than that.

If you need a more formal answer, please describe the problem, as well as the algorithm being used, with more details.