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: knapsack algorithm in python
# a dynamic approach
# Returns the maximum value that can be stored by the bag
def knapSack(W, wt, val, n):
K = [[0 for x in range(W + 1)] for x in range(n + 1)]
#Table in bottom up manner
for i in range(n + 1):
for w in range(W + 1):
if i == 0 or w == 0:
K[i][w] = 0
elif wt[i-1] <= w:
K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])
else:
K[i][w] = K[i-1][w]
return K[n][W]
#Main
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print(knapSack(W, wt, val, n))
Example 3: knapsack problem
#include <bits/stdc++.h>
using namespace std;
int n, w;
int main() {
cin >> n >> w;
vector<long long> dp(w + 1, 0);
for (int i = 0; i < n; ++i) {
int value, cost;
cin >> value >> cost;
for (int j = w; j >= cost; --j)
dp[j] = max(dp[j], value + dp[j - cost]);
}
cout << dp[w];
}
Example 4: knapsack
#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
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){
if(x>=n or b<=0)return 0;
if(dp[x][b]!=-1)return dp[x][b];
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(){
ios_base::sync_with_stdio(0);cin.tie(0);
cin>>n>>s;
a.resize(n);
for(int i=0;i<n;i++)
cin>>a[i].first;
for(int i=0;i<n;i++)
cin>>a[i].second;
ini();
cout<<f(0,s);
return 0;
}