0/1 knapsack problem code example

Example 1: greedy knapsack

def greedy_knapsack(values,weights,capacity):
    n = len(values)
    def score(i) : return values[i]/weights[i]
    items = sorted(range(n)  , key=score , reverse = True)
    sel, value,weight = [],0,0
    for i in items:
        if weight +weights[i] <= capacity:
            sel += [i]
            weight += weights[i]
            value += values [i]
    return sel, value, weight


weights = [4,9,10,20,2,1]
values = [400,1800,3500,4000,1000,200]
capacity = 20

print(greedy_knapsack(values,weights,capacity))

Example 2: python 0-1 kanpsack

#Returns the maximum value that can be stored by the bag

def knapSack(W, wt, val, n):
   # initial conditions
   if n == 0 or W == 0 :
      return 0
   # If weight is higher than capacity then it is not included
   if (wt[n-1] > W):
      return knapSack(W, wt, val, n-1)
   # return either nth item being included or not
   else:
      return max(val[n-1] + knapSack(W-wt[n-1], wt, val, n-1),
         knapSack(W, wt, val, n-1))
# To test above function
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print (knapSack(W, wt, val, n))

Example 3: knapsack

#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
//dp table is full of zeros
int n,s,dp[1002][1002];
void ini(){
    for(int i=0;i<1002;i++)
        for(int j=0;j<1002;j++)
            dp[i][j]=-1;
}
int f(int x,int b){
	//base solution
	if(x>=n or b<=0)return 0;
	//if we calculate this before, we just return the answer (value diferente of 0)
	if(dp[x][b]!=-1)return dp[x][b];
	//calculate de answer for x (position) and b(empty space in knapsack)
	//we get max between take it or not and element, this gonna calculate all the
	//posible combinations, with dp we won't calculate what is already calculated.
	return dp[x][b]=max(f(x+1,b),b-a[x].second>=0?f(x+1,b-a[x].second)+a[x].first:INT_MIN);
}
int main(){
	//fast scan and print
	ios_base::sync_with_stdio(0);cin.tie(0);
	//we obtain quantity of elements and size of knapsack
	cin>>n>>s;
	a.resize(n);
	//we get value of elements
	for(int i=0;i<n;i++)
		cin>>a[i].first;
	//we get size of elements
	for(int i=0;i<n;i++)
		cin>>a[i].second;
	//initialize dp table
	ini();
	//print answer
	cout<<f(0,s);
	return 0;
}