Talking about Packet backpack

Packet backpack, a condition evolved from the base backpack.


The specific question is this. The specific question is this. A backpack with a capacity of V, and several sets of items, each containing a number of items, which are different, and volume W and value p are different. Objects in the group conflict. The maximum value can be obtained in the case of no more than v.


At first glance it seems very difficult to look, in fact, carefully think about it is very simple, this problem can be solved with 01 knapsack. For packet packs, can think so: although divided into many groups, but can only choose one, or not, this and 01 backpack is the same, that is, for each of the 01 backpack unique items, the corresponding packet pack is each group to select an item, so look, is exactly 01 knapsack problem.


Here is the equation of state:
for (int i = 1; I <= z; i++)
for (int j = V; J >= 1; j--)
for (int k = 1; k <= N; k++)
DP[J] = max (Dp[j-w[i][k]] + p[i][k], dp[j]);


where z is the number of packets, V is the backpack volume, n is the number of items per group, W and P are the volume and value of the first group K items.


Why do you write that? Think about the equation of state for the 01 backpack, which is the same as this outer two-layer loop, not the same is added to the inside of the layer, this cycle is to traverse each group of items used, for dp[] Each state is a cycle of each group of items only to the next one, so no effect on the back, It is guaranteed that there is only one item in each group.