Never been to DZone Snippets before?

Snippets is a public source code repository. Easily build up your personal collection of code snippets, categorize them with tags / keywords, and share them with the world

« Newer Snippets
Older Snippets »
Showing 1-2 of 2 total  RSS 

The 0-1 Knapsack Problem in C

This is a dynamic-programming algorithm implementation for solving the the 0-1 Knapsack Problem in C.

Further explanation is given here.

#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;
}

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 <stdio.h>

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;
}
« Newer Snippets
Older Snippets »
Showing 1-2 of 2 total  RSS