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This is a dynamic-programming algorithm implementation for solving the <a href="http://en.wikipedia.org/wiki/Knapsack_problem">the 0-1 Knapsack Problem</a> in C. Further explanation is given <a href="http://compprog.wordpress.com/2007/11/20/the-0-1-knapsack-problem/">here</a>.

#include <stdio.h> #define MAXWEIGHT 100 int n = 3; /* The number of objects */ int c[10] = {8, 6, 4}; /* c[i] is the *COST* of the ith object; i.e. what YOU PAY to take the object */ int v[10] = {16, 10, 7}; /* v[i] is the *VALUE* of the ith object; i.e. what YOU GET for taking the object */ int W = 10; /* The maximum weight you can take */ void fill_sack() { int a[MAXWEIGHT]; /* a[i] holds the maximum value that can be obtained using at most i weight */ int last_added[MAXWEIGHT]; /* I use this to calculate which object were added */ int i, j; int aux; for (i = 0; i <= W; ++i) { a[i] = 0; last_added[i] = -1; } a[0] = 0; for (i = 1; i <= W; ++i) for (j = 0; j < n; ++j) if ((c[j] <= i) && (a[i] < a[i - c[j]] + v[j])) { a[i] = a[i - c[j]] + v[j]; last_added[i] = j; } for (i = 0; i <= W; ++i) if (last_added[i] != -1) printf("Weight %d; Benefit: %d; To reach this weight I added object %d (%d$ %dKg) to weight %d.\n", i, a[i], last_added[i] + 1, v[last_added[i]], c[last_added[i]], i - c[last_added[i]]); else printf("Weight %d; Benefit: 0; Can't reach this exact weight.\n", i); printf("---\n"); aux = W; while ((aux > 0) && (last_added[aux] != -1)) { printf("Added object %d (%d$ %dKg). Space left: %d\n", last_added[aux] + 1, v[last_added[aux]], c[last_added[aux]], aux - c[last_added[aux]]); aux -= c[last_added[aux]]; } printf("Total value added: %d$\n", a[W]); } int main(int argc, char *argv[]) { fill_sack(); return 0; }