0-1 knapsack problem (Quick Start to Dynamic Planning)

9. Questions about backpacks

There are n items and a backpack with a capacity of v. The cost of the I-th item is C [I], and the value is W [I]. Solving which items are loaded into a backpack can maximize the total value.

Basic Ideas

This is the most basic problem with a backpack. It features that each item has only one item, and you can choose to put it or not.

Define the state with a subproblem: that is, F [I] [v] indicates the maximum value that a backpack with a capacity of V can obtain when the first I item is placed. The state transition equation is:

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

This equation is very important. Basically all the equations related to the backpack are derived from it. Therefore, it is necessary to explain in detail: "Put the first I items into a backpack with a capacity of V, if you only consider the I-th item Policy (put or not put), then it can be converted into a problem that only involves the previous I-1 items. If I items are not put, then the problem is converted to "pre-i-1 items into the capacity of V backpack", the value is f [I-1] [v]; if I items are placed, the problem is converted to "the previous I-1 items are placed in the backpack with the remaining capacity V-C [I ", the greatest value that can be obtained at this time is f [I-1] [V-C [I] plus the value W [I] obtained by placing the I item.

The source code to paste below.

Int V [200] [200]; // The maximum value int max (int A, int B) obtained by loading the first I item into a backpack with a capacity of J) {if (a> = B) return a; else return B;} int knapsack (INT N, int W [], int V [], int X [], int C) {int I, j; for (I = 0; I <= N; I ++) V [I] [0] = 0; For (j = 0; j <= C; j ++) V [0] [J] = 0; for (I = 0; I <= n-1; I ++) for (j = 0; j <= C; j ++) if (j <W [I]) V [I] [J] = V [I-1] [J]; else V [I] [J] = max (V [I-1] [J], V [I-1] [J-W [I] + V [I]); j = C; for (I = n-1; I> = 0; I --) {If (V [I] [J]> V [I-1] [J]) {x [I] = 1; j = J-W [I];} else X [I] = 0;} printf ("the selected item is \ n "); for (I = 0; I <n; I ++) printf ("% d", X [I]); printf ("\ n "); return V [n-1] [c];} void main () {int s; // obtain the maximum value of int W [15]; // item weight int V [15]; // item value int X [15]; // item selection status int N, I; int C; // maximum backpack capacity n = 5; printf ("Enter the maximum backpack capacity: \ n"); scanf ("% d", & C ); printf ("input item count: \ n"); scanf ("% d", & N); printf ("Enter the item weight separately: \ n "); for (I = 0; I <n; I ++) scanf ("% d", & W [I]); printf ("Enter the value of the item separately: \ n "); for (I = 0; I <n; I ++) scanf (" % d ", & V [I]); s = knapsack (n, w, V, X, c); printf ("Maximum item value: \ n"); printf ("% d \ n", S );}