Solving steps:
1) model
2) Find constraints : only three commodity, a weight of the backpack 10
3) to find recurrence relations
V (i): Value
W is ( I ) : Weight
V ( i, J ): the current capacity of the backpack J , before the i th best combination value corresponding to the article
For current product there are two cases:
① current product weight is greater than the weight of the backpack remaining, not put into. Then V ( I-. 1, J ) = V ( I, J ) ;
② current product weight is less than the remaining weight backpack, but not necessarily installed can reach the optimal solution, and the refill is not installed between a selected. Max {v (i-1, j), v (i, jw (i)) + v (i)}
Wherein V (I-. 1, J) : indicates not installed; V (I, JW (I)) + V (I) represents installed, backpack remaining weight reduction w (i), but the value increases v (i). It follows the recurrence relation:
j(i)>w(i): v(i,j)=v(i-1,j)
J(i)<w(i): max{v(i-1,j),v(i-1,j-w(i))+v(i)}
Note: Why put the case can go to solving such as dynamic programming there is a principle of optimality, v (i-1, JW (i)) is a condition caused by earlier decisions, is simply the present article can be put to, but I need to make a comparison, the previous comparison was already out of the optimal solution will not be because the backpack adds new weight has changed, from a bag and go in and hold the two cases selected maximum value. Behind decisions must constitute an optimal strategy. Two cases were compared, the best results.