In the 0-1 Knapsack problem, we are given a set of items with individual weights and profits and a container of fixed capacity (the knapsack), and are required to compute a loading of the knapsack with items such that the total profit is maximised. We only have 1 of each item, so there is either 0 or 1 of each item in in the knapsack, hence the 0-1 in the name of the problem.

The 0-1 knapsack problem is NP-hard, but can be solved quite efficiently using backtracking. Below is a backtracking implementation in C. The function knapsack() takes arrays of weights, and profits, their size, the capacity, and the address of a pointer through which the solution array is returned. It returns the profit of the best knapsack. The profit density of each item (weight / profit) is first calculated, and the knapsack is filled in order of decreasing profit density. A bounding function is used to determine the upper bound on the profit achievable with the current knapsack so as to reduce the search space.

#include <stdlib.h>
#include <string.h>

/* A knapsack item */
typedef struct {
    unsigned int id;
    double weight;
    double profit;
    double profit_density;
} item;

/* Compare items by lesser profit density */
static int compare_items(const item *item1, const item *item2)
{
    if (item1->profit_density > item2->profit_density) {
        return -1;
    }
    if (item1->profit_density < item2->profit_density) {
        return 1;
    }
    return 0;
}

/* Bounding function */
static double profit_bound(const item *items, size_t n, double capacity,
        double current_weight, double current_profit,
        unsigned int level)
{
    double remaining_capacity = capacity - current_weight;
    double bound = current_profit;
    unsigned int lvl = level;
    /* Fill in order of decreasing profit density */
    while (lvl < n &&
        items[lvl].weight <= remaining_capacity) 
    {
        remaining_capacity -= items[lvl].weight;
        bound += items[lvl].profit;
        lvl++;
    }
    /* Pretend we can take a fraction of the next object */
    if (lvl < n) {
        bound += items[lvl].profit_density
            * remaining_capacity;
    }
    return bound;
}

static void knapsack_recursive(const item *items, size_t n, double capacity,
        unsigned int *current_knapsack, double *current_weight, double *current_profit,
        unsigned int *max_knapsack, double *max_profit, unsigned int level)
{   
    if (level == n) {
        /* Found a new max knapsack */
        *max_profit = *current_profit;
        memcpy(max_knapsack, current_knapsack, n * sizeof(unsigned int));
        return;
    }
    if (*current_weight + items[level].weight <= capacity)
    {   /* Try adding this item */
        *current_weight += items[level].weight;
        *current_profit += items[level].profit;
        current_knapsack[items[level].id] = 1;
        knapsack_recursive(items, n, capacity, current_knapsack, current_weight, 
                current_profit, max_knapsack, max_profit, level + 1);
        *current_weight -= items[level].weight;
        *current_profit -= items[level].profit;
        current_knapsack[items[level].id] = 0;
    }
    if (profit_bound(items, n, capacity, *current_weight, 
                *current_profit, level + 1) > *max_profit) {
        /* Still promising */
        knapsack_recursive(items, n, capacity, current_knapsack, current_weight, 
                current_profit, max_knapsack, max_profit, level + 1);
    }
}

double knapsack(const double *weights, const double *profits, 
        size_t n, double capacity, unsigned int **max_knapsack)
{
    double current_weight = 0.0;
    double current_profit = 0.0;
    double max_profit = 0.0;
    unsigned int i;
    item *items  = malloc(n * sizeof(item));
    unsigned int *current_knapsack = calloc(n, sizeof(unsigned int));
    *max_knapsack = malloc(n * sizeof(unsigned int));
    if (!(items && current_knapsack && *max_knapsack)) {
        free(items);
        free(current_knapsack);
        free(*max_knapsack);
        *max_knapsack = NULL;
        return 0;
    }
    /* Populate the array of items */
    for (i = 0; i < n; i++) {
        items[i].id = i;
        items[i].weight = weights[i];
        items[i].profit = profits[i];
        items[i].profit_density = profits[i] / weights[i];
    }
    /* Sort into decreasing density order */
    qsort(items, n, sizeof(item),
            (int (*)(const void *, const void *))compare_items);
    knapsack_recursive(items, n, capacity, current_knapsack, &current_weight,
            &current_profit, *max_knapsack, &max_profit, 0);
    free(items);
    free(current_knapsack);
    return max_profit;
}

Here is an example program:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    double weights[] = {3, 5, 2, 1};
    double profits[] = {9, 10, 7, 4};
    const size_t n = sizeof(profits) / sizeof(profits[0]);
    const double capacity = 7;
    unsigned int *max_knapsack;
    double max_profit = knapsack(weights, profits, n, capacity, &max_knapsack);
    unsigned int i;
    printf("Profit: %.2f\n", max_profit);
    printf("Knapsack contains:\n");
    for (i = 0; i < n; i++) {
        if (max_knapsack[i] == 1) {
            printf("Item %u with weight %.2f and profit %.2f\n", i, weights[i], profits[i]); 
        }
    }
    free(max_knapsack);
    return 0;
}

The output:

Profit: 20.00
Knapsack contains:
Item 0 with weight 3.00 and profit 9.00
Item 2 with weight 2.00 and profit 7.00
Item 3 with weight 1.00 and profit 4.00