The Fractional Knapsack Problem In C

This is the classic Greedy algorithm implementation for solving the Fractional Knapsack Problem in C.

Further explanations here


#include 

int n = 5; /* The number of objects */
int c[10] = {12, 1, 2, 1, 4}; /* c[i] is the *COST* of the ith object; i.e. what
				YOU PAY to take the object */
int v[10] = {4, 2, 2, 1, 10}; /* v[i] is the *VALUE* of the ith object; i.e.
				what YOU GET for taking the object */
int W = 15; /* The maximum weight you can take */

void simple_fill() {
	int cur_w;
	float tot_v;
	int i, maxi;
	int used[10];

	for (i = 0; i < n; ++i)
		used[i] = 0; /* I have not used the ith object yet */

	cur_w = W;
	while (cur_w > 0) { /* while there's still room*/
		/* Find the best object */
		maxi = -1;
		for (i = 0; i < n; ++i)
			if ((used[i] == 0) &&
				((maxi == -1) || ((float)v[i]/c[i] > (float)v[maxi]/c[maxi])))
				maxi = i;

		used[maxi] = 1; /* mark the maxi-th object as used */
		cur_w -= c[maxi]; /* with the object in the bag, I can carry less */
		tot_v += v[maxi];
		if (cur_w >= 0)
			printf("Added object %d (%d$, %dKg) completly in the bag. Space left: %d.\n", maxi + 1, v[maxi], c[maxi], cur_w);
		else {
			printf("Added %d%% (%d$, %dKg) of object %d in the bag.\n", (int)((1 + (float)cur_w/c[maxi]) * 100), v[maxi], c[maxi], maxi + 1);
			tot_v -= v[maxi];
			tot_v += (1 + (float)cur_w/c[maxi]) * v[maxi];
		}
	}

	printf("Filled the bag with objects worth %.2f$.\n", tot_v);
}

int main(int argc, char *argv[]) {
	simple_fill();

	return 0;
}