Implement 0/1 Knapsack problem using Dynamic Programming

				
#include<stdio.h>
#include<conio.h>
int w[10],p[10],v[10][10],n,i,j,cap,x[10]={0};
int max(int i,int j)
{
return ((i>j)?i:j);
}
int knap(int i,int j)
{
int value; if(v[i][j]<0)
{
if(j<w[i])
value=knap(i-1,j);
else
value=max(knap(i-1,j),p[i]+knap(i-1,j-w[i]));
v[i][j]=value;
}
return(v[i][j]);
}
void main()
{
int profit,count=0;
clrscr();
printf("\nEnter the number of elements\n");
scanf("%d",&n);
printf("Enter the profit and weights of the elements\n");
for(i=1;i<=n;i++)
{
printf("For item no %d\n",i);
scanf("%d%d",&p[i],&w[i]);
}
printf("\nEnter the capacity \n");
scanf("%d",&cap);
for(i=0;i<=n;i++)
for(j=0;j<=cap;j++)
if((i==0)||(j==0))
v[i][j]=0;
else
v[i][j]=-1;
profit=knap(n,cap);
i=n;
j=cap;
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 included are\n");
printf("Sl.no\tweight\tprofit\n");
for(i=1;i<=n;i++)
if(x[i]) printf("%d\t%d\t%d\n",++count,w[i],p[i]);
printf("Total profit = %d\n",profit);
getch();
}