Dynamic Programming 0-1 knapsack problem

Dynamic Programming 0-1 knapsack problem

? Description: N items and a backpack are given. The Weight of item I is wi, its value is VI, and the size of the backpack is C. How should I select the items to be loaded into the backpack to maximize the total value of the items in the backpack ?? For an item, either it is packed into a backpack or it is not loaded. Therefore, the loading status of an item can be 0 and 1. We set the loading status of item I to XI, XI, (0, 1). This problem is called the 0-11 backpack problem. Process Analysis

Data: number of items n = 5, item weight W [N] = {,}, item value V [N] = },

(Set 0th bits to 0, which is easy to unify with The subscripts. It is not particularly useful and cannot be processed in this way .) Total weight c = 10.

? The maximum size of the backpack is 10, so when the array m size is set, the row and column values can be set to 6 and 11, then, for M (I, j), it indicates that the optional items are I... When the n-pack capacity is J (total weight), the maximum value of the items in the backpack is.

 

 

 

The source code is as follows:

# Include <stdio. h> # include <stdlib. h> # include <iostream> # include <queue> # include <climits> # include <cstring> using namespace STD; const int c = 10; // The capacity of the backpack const int W [] = {0, 2, 6, 5, 4}; // the weight of the item, which is not used at position 0. Const int V [] = {0, 6, 3, 5, 4, 6}; // the corresponding item to be added, position 0 is left blank. Const int n = sizeof (w)/sizeof (W [0])-1; // n is the number of items int X [n + 1]; void package0_1 (int m [] [11], const int W [], const int V [], const int N) // n indicates the number of items {// The sequence from bottom to top is used to set the value of M [I] [J] // put W [N] For (Int J = 0; j <= C; j ++) if (j <W [N]) m [N] [J] = 0; // J is less than W [N], set the corresponding value to 0. Otherwise, else M [N] [J] = V [N] can be placed. // you can place the remaining n-1 items. Int I; for (I = n-1; I> = 1; I --) for (Int J = 0; j <= C; j ++) if (j <W [I]) m [I] [J] = m [I + 1] [J]; // assuming j <W [I, the current position cannot be placed. It is equal to the value of the previous position. // Otherwise, it will be larger than the value after the marker is placed or not placed. Else M [I] [J] = m [I + 1] [J]> M [I + 1] [J-W [I] + V [I]? M [I + 1] [J]: M [I + 1] [J-W [I] + V [I];} void answer (int m [] [11], const int N) {Int J = C; int I; for (I = 1; I <= n-1; I ++) if (M [I] [J] = m [I + 1] [J]) x [I] = 0; else {x [I] = 1; j = J-W [I];} X [N] = m [I] [J]? 1: 0;} int main () {int M [6] [11] = {0}; package0_1 (M, W, V, N ); for (INT I = 0; I <= 5; I ++) {for (Int J = 0; j <= 10; j ++) printf ("% 2D ", M [I] [J]); cout <Endl;} answer (m, n); cout <"the best answer is: \ n "; for (INT I = 1; I <= 5; I ++) cout <X [I] <""; System ("pause"); Return 0 ;}

 

Dynamic Programming 0-1 knapsack problem