Fractional Knapsack Problem in C

This is a C Program to solve fractional knapsack problem. The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a mass and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.

Here is the source code of the C program to solve fractional knapsack problem. The C program is successfully compiled and run on a Linux system. The program output is also shown below.

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#include <stdio.h>
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int n = 5; /* The number of objects */
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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 */

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void simple_fill() {
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    int cur_w;
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    float tot_v;
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    int i, maxi;
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    int used[10];
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    for (i = 0; i < n; ++i)
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        used[i] = 0; /* I have not used the ith object yet */

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    cur_w = W;
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    while (cur_w > 0) { /* while there's still room*/

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        /* Find the best object */
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        maxi = -1;
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        for (i = 0; i < n; ++i)
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            if ((used[i] == 0) &&
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                ((maxi == -1) || ((float)v[i]/c[i] > (float)v[maxi]/c[maxi])))

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                maxi = i;
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        used[maxi] = 1; /* mark the maxi-th object as used */

        cur_w -= c[maxi]; /* with the object in the bag, I can carry less */

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        tot_v += v[maxi];
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        if (cur_w >= 0)
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            printf("Added object %d (%d$, %dKg) completely in the bag. Space left: %d.\n", maxi + 1, v[maxi], c[maxi], cur_w);

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        else {
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            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);

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            tot_v -= v[maxi];
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            tot_v += (1 + (float)cur_w/c[maxi]) * v[maxi];

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

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}
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int main(int argc, char *argv[]) {
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    simple_fill();
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    return 0;
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}
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$ gcc fractional_knapsack.c -o fractional_knapsack
$ ./fractional_knapsack
 
Added object 5 (10$, 4Kg) completely in the bag. Space left: 11.
Added object 2 (2$, 1Kg) completely in the bag. Space left: 10.
Added object 3 (2$, 2Kg) completely in the bag. Space left: 8.
Added object 4 (1$, 1Kg) completely in the bag. Space left: 7.
Added 58% (4$, 12Kg) of object 1 in the bag.
Filled the bag with objects worth 17.33$.