Question

There is a simple dp solution of this problem:

#include <vector>
#include <limits>

int knapsack2(const std::vector<int>& wts, const std::vector<int>& cost, int W)
{
    size_t n = wts.size();
    std::vector<std::vector<int> > dp(W + 1, std::vector<int>(n+1, 0));
    for (size_t j = 1; j <= n; j++) {
        for (int w = 1; w <= W; w++) {
           if (wts[j-1] <= w) {
               dp[w][j] = std::max(dp[w][j - 1], dp[w - wts[j-1]][j - 1] + cost[j-1]);
           }
           else {
                dp[w][j] = dp[w][j - 1];
           }
        }
    }
    return dp[W][n];
}

But how can I know what objects were taken?

Answer 1

Make the second vector “prev” with the same dimensions as dp.

When you choose the best option for dp[w][j], write coordinates of previous cell to prev[w][j]

When the job of dp is done, walk from prev[W][n] back to get all the used cell indexes