how to do knapsack problem code example
Example 1: knapsack problem
// memory efficient and iterative approach to the knapsack problem
using namespace std;
// n is the number of items
// w is the knapsack's capacity
int n, w;
int main() {
/*
input format:
n w
value_1 cost_1
value_2 cost_2
.
.
value_n cost_n
*/
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]);
}
// the answer is dp[w]
cout << dp[w];
}
Example 2: knapsack
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;
}