Created
September 11, 2013 14:36
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KnapSack Problem implemented in C using Recursion
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#include <stdio.h> | |
#include <stdlib.h> | |
int w[10], p[10], v[10][10], n, i, j, capacity, x[10] = {0}; | |
int max(int i, int j) { | |
return ((i > j) ? i : j); | |
} | |
int KnapSack(int i, int j) { | |
int value; | |
if (v[i][j] < 0) { | |
if (j < w[i]) | |
value = KnapSack(i - 1, j); | |
else | |
value = max(KnapSack(i - 1, j), p[i] + KnapSack(i - 1, j - w[i])); | |
v[i][j] = value; | |
} | |
return (v[i][j]); | |
} | |
int main(int argc, char** argv) { | |
int profit, count = 0; | |
printf("\nEnter the number of elements : "); | |
scanf("%d", &n); | |
printf("\nEnter the profit and weights of the elements\n"); | |
for (i = 1; i <= n; i++) { | |
printf("Item no : %d\n", i); | |
scanf("%d %d", &p[i], &w[i]); | |
} | |
printf("\nEnter the capacity \n"); | |
scanf("%d", &capacity); | |
for (i = 0; i <= n; i++) | |
for (j = 0; j <= capacity; j++) | |
if ((i == 0) || (j == 0)) | |
v[i][j] = 0; | |
else | |
v[i][j] = -1; | |
profit = KnapSack(n, capacity); | |
i = n; | |
j = capacity; | |
while (j != 0 && i != 0) { | |
if (v[i][j] != v[i - 1][j]) { | |
x[i] = 1; | |
j = j - w[i]; | |
i--; | |
} else | |
i--; | |
} | |
printf("Items in the KnapSack are : \n\n"); | |
printf("Sl.no \t weight \t profit\n"); | |
printf("\n----------------------------------------\n"); | |
for (i = 1; i <= n; i++) | |
if (x[i]) | |
printf("%d \t %d \t\t %d\n", ++count, w[i], p[i]); | |
printf("Total profit = %d\n", profit); | |
return (EXIT_SUCCESS); | |
} | |
/* | |
Output:- | |
Enter the number of elements : 3 | |
Enter the profit and weights of the elements | |
Item no : 1 | |
4 6 | |
Item no : 2 | |
1 2 | |
Item no : 3 | |
7 3 | |
Enter the capacity | |
7 | |
Items in the KnapSack are : | |
Sl.no weight profit | |
---------------------------------------- | |
1 2 1 | |
2 3 7 | |
Total profit = 8 | |
*/ |
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