Retrieval of items from a knapsack using dynamic programming - c++

Question

I'm new to dynamic programming and have tried my first DP problem. The problem statement is
Given a knapsack of size C, and n items of sizes s[] with values v[], maximize the capacity of the items which can be put in the knapsack. An item may be repeated any number of times. (Duplicate items are allowed).
Although I was able to formulate the recurrence relation and create the DP table, and eventually get the maximum value that can be put in the knapsack, I am not able to device a method to retrieve which values have to be selected to get the required sum.
Here is my solution:
#include <iostream>
#include <cstdlib>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
int s[] = { 1, 3, 4, 5, 2, 7, 8 , 10};
int v[] = { 34, 45, 23, 78, 33, 5, 7 , 1};
int n = ( (sizeof(s)) / (sizeof(s[0])) );
vector<int> backtrack;
int C = 15;
int pos;
int m[20];
m[0] = 0;
int mx = 0;
for ( int j = 1; j <= C; j++) {
mx = 0;
m[j] = m[j-1];
pos = j-1;
for ( int i = 0; i < n; i++) {
mx = m[i-s[i]] + v[i];
if ( mx > m[i] ) {
m[i] = mx;
pos = i - s[j];
}
}
backtrack.push_back(pos);
}
cout << m[C] << endl<<endl;
for ( int i = 0; i < backtrack.size(); i++) {
cout << s[backtrack[i]] <<endl;
}
return 0;
}
In my solution, I've attempted to store the positions of the maximum value item selcted in a vector, and eventually print them. However this does not seem to give me the correct solution.
Running the program produces:
79
2
3
0
5
2
7
8
10
34
45
23
78
33
5
7
It is obvious from the output that the numbers in the output cant be the sizes of the items selected as there there no item of size 0 as shown in the output.
I hope that you will help me find the error in my logic or implementation. Thanks.

Answer 1

You are following the greedy approach. It's pretty clever , but it is heuristic. I will not give you the correct code, as it might be a homework, but the recursive function knapsack would look like this:
knapsack(C): maximum profit achivable using a knapsack of Capacity C
knapsack(C) = max { knapsack(C-w[i]) + v[i] } for all w[i] <= C
knapsack(0) = 0
In code:
dp(0) = 0;
for i = 1 to C
dp(i) = -INF;
for k = i-1 downto 0
if w[k] < i then
dp(i) = max{dp(i-w[k]) + v[k], dp(i)};
print dp(Capacity);