Multiplepack coming ~ ^. ^

Multiple backpacks:

Basic Ideas:

Let's take a look at an example: there are n items and a backpack with a capacity of v. A maximum of N [I] items are available for the I-th item. The cost per item is C [I] and the value is W [I]. Solving which items are loaded into a backpack can make the total cost of these items not exceed the capacity of the backpack, and the total value is the largest. This type of backpack is characterized by a limited number of items.

It is similar to a full backpack. The basic equation only needs to slightly change the equation of the complete backpack problem, because there is n [I] + 1 strategy for the I-th item: Take 0 pieces, 1 piece... n [I] pieces. If f [I] [v] indicates that the first I items are placed in the maximum weight of a backpack with a capacity of V, there is a state transition equation: f [I] [v] = max {f [I-1] [V-K * C [I] + K * W [I] | 0 <= k <= N [i]} the range of K in the complete backpack is (0 <= k <= V/C [I]). complexity is O (V * Σ N [I]),

The difference between the Code and the full backpack is only in the internal loop

1	for(k = 1; k <= j/weight[i]; ++k)

Change

1	for(k = 1; k <=n[i] && k<=j/weight[i]; ++k)

Binary splitting of multiple backpacks

In the same way as a backpack, the binary idea is used to split N [I] items into several items, the purpose is that any number in 0-N [I] can be exchanged with these items. In addition, policies that exceed n [I] parts are not allowed.

The method is to divide item I into several items, each of which has a coefficient. The cost and value of this item are the original cost and value multiplied by this coefficient, make these coefficients respectively 1, 2, 4 ,..., 2 ^ (k-1), N [I]-2 ^ k + 1, and K satisfies the maximum integer of N [I]-2 ^ k + 1> 0. For example, if n [I] = 13, the item is split into four items with a coefficient of 1, 2, 4, and 6. The coefficient of the divided items is n [I], indicating that the I-th item cannot be obtained more than N [I] items. In addition, this method ensures that each integer between 0 .. n [I] can be expressed by the sum of several coefficients.

1 # include <iostream> 2 using namespace STD; 3 4/* multi-backpack binary splitting 5 * time complexity greater than O (N * V) 6 * space complexity O (N * V) 7 * Set v <= 200 n <= 10, total number of items after splitting <50 8 * each item has log n [I] Status 9 */10 11 int maxv [201]; 12 INT weight [50]; /* record the weight of the object after splitting */13 int value [50];/* record the value of the object after splitting */14 int V, N; 15 16 void main () 17 {18 int I, j; 19 scanf ("% d", & V, & N); 20 int weig, Val, num; 21 int COUNT = 0; 22 23 for (I = 0; I <N; ++ I) 24 {25 scanf ("% d", & weig, & Val, & num); 26 27 for (j = 1; j <= num; j <= 1) // binary split 28 {29 weight [count] = J * weig; 30 value [count ++] = J * val; 31 num-= J; 32} 33 If (Num> 0) 34 {35 weight [count] = num * weig; 36 value [count ++] = num * val; 37} 38} 39 for (I = 0; I <count; ++ I) // use 01 backpack 40 {41 for (j = V; j> = weight [I]; -- j) 42 {43 int TMP = maxv [J-weight [I] + value [I]; 44 maxv [J] = maxv [J]> T MP? Maxv [J]: TMP; 45} 46} 47 printf ("% d", maxv [v]); 48} 49 50/* 51 [input sample] 52 4 2053 3 9 354 5 9 155 9 4 256 8 1 357 [output sample] 58 4759 */

The following describes how o (log amount) processes items in multiple backpacks. amount indicates the number of items:

 1 void multiplepack(int cost, int weight, int amount) 2 { 3     if(cost*amount>=V) 4         cmpletepack(cost,weight); 5     else 6     { 7         int k=1; 8         while(k<amount) 9         {10             zeroonepack(k*cost,k*weight);11             amount-= k;12             k=k*2;13         }14         zeroonepack(amount*cost,amount*weight);15     }16 }

Conclusion ~ I can't answer questions when I want to go home ~ Xiaoxu ~ But I am still insisting on updating my blog every day ~ *. *