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The 0-1 Knapsack Problem In C

11.20.2007
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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;
}