0-1 knapsack problem-Dynamic Programming

0-1 knapsack problem-Dynamic Programming
Problem description: n items and a backpack are given. The Weight of item I is w [I], the value is v [I], and the size of the backpack is C. Q: How should I select the items to be loaded into the backpack to maximize the total value of the items in the backpack?
Analysis: For an item, either a backpack or no package is loaded. Therefore, the loading status of an item can be 0 or 1. Set the loading status of item I to xi, xi, (0, 1). This problem is called the 0-1 backpack problem.
Data: number of items n = 5, item weight w [5] = {2, 6, 5, 4}, item value v [5] = {6, 3, 5, 4, 6 }, total weight c = 10. The maximum size of the backpack is 10, so in the array m size, you can set the row and column values to 5 and 10, then, for m (I, j) it indicates the maximum value of items in the backpack when the optional items are I to n and the size of the backpack is j (total weight. The following table shows the process of solving the 0-1 knapsack problem by using the dynamic programming method:
Vc + jrMTH1ta94bn7tcS82yYjMjA1NDA7tPO + signature + PHByZSBjbGFzcz0 = "brush: java;"> # include Const int c = 10; // capacity of the backpack const int w [] = {, 4}; // item weight const int v [] = {6, 3, 5, 4, 6}; // const int n = sizeof (w)/sizeof (w [0])-1; // n indicates the number of items int x [n]; void package0_1 (int m [] [10], const int w [], const int v [], const int n) // n indicates the number of items {int I, j; /********************************** place the last item separately * * ******************************/for (j = 1; j <= c; j ++) {if (j <w [n])/* backpack capacity <最后一个物品的重量时* m[n][j-1]="v[n];" else *背包能够放下最后一个物品时* } *********************************放置前n-1个物品********************************* for(i="n-1;" i> = 0; I --) for (j = 1; j <= c; j ++) {if (j <w [I]) m [I] [J-1] = m [I + 1] [J-1]; // If j <w [I], the current position cannot be placed, it is equal to the value of else at the previous position. Otherwise, it compares whether the placed value is large or not, choose m [I] [J-1] = m [I + 1] [J-1]> m [I + 1] [j-1-w [I] + v [I]? M [I + 1] [J-1]: m [I + 1] [j-1-w [I] + v [I];} void answer (int m [] [10], const int n) {int j = C-1;/* I = 0, j = the maximum value of storing a backpack with c x/int I; for (I = 0; I <n; I ++) if (m [I] [j] = m [I + 1] [j]) x [I] = 0; else {x [I] = 1; /* if the current item is in a backpack */j = j-w [I];/* re-calculate the remaining size of the backpack, in order to calculate the trade-off between other items during the optimization */} I-= 1; x [n] = m [I] [j]? 1: 0;} int main () {int m [6] [10] = {0}; int I, j; package0_1 (m, w, v, n ); for (I = 0; I <= 4; I ++) {for (j = 0; j <10; j ++) printf ("% 3d ", m [I] [j]); printf ("\ n");} answer (m, n); printf ("The best answer is: \ n "); for (I = 0; I <5; I ++) printf ("% d", x [I]); printf ("\ n"); return 0 ;}

The program running result is as follows: