Road Optimization.cpp

// https://codeforces.com/contest/1625/problem/C
// Dynamic programming
//  299143BT
#include<bits/stdc++.h>
#define int long long int
#define MOD 1000000007
#define pi 3.1415926535
#define fi first
#define se second
#define vll vector<int>
#define vpll vector<pair <int,int> >
#define vvll vector< vll >
#define str string
#define umap unordered_map
#define uset unordered_set
#define rep(i,n) for(int i=0;i<n;i++)
#define repA(i,s,n) for(int i=s;i<=n;i++)
#define repD(i,s,n) for(int i=s;i>=n;i--)
#define pb push_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define elif else if
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define ALL(v) v.begin(),v.end()
#define sortA(v) sort(ALL(v))
#define sortD(v) sort(ALL(v),greater <int> ())
#define print(v) for(auto itr:v) cout<<itr<<" ";
#define sz size()
#define nl cout<<"\n";
using namespace std;
int factorial(int n){return (n==1||n==0)?1:n*factorial(n-1);}
int power(int x,int y){int res=1;while(y>0){if(y&1)res=res*x;y=y>>1;x=x*x;}return res;}
int powerMod(int x,int y,int m){if(y==0)return 1;int p=powerMod(x,y/2,m)%m;p=(p*p)%m;return (y%2==0)?p:(x*p)%m;}
int modInverse(int a,int m){return powerMod(a,m-2,m);}
int gcd(int a,int b){if(a==0){return b;}else if(b==0){return a;}else{if(a>b){return gcd(a%b,b);}else{return gcd(b%a,a);}}}
int ceil_div(int a,int b){return (a/b)+((a%b)>0);}

int dp[509][509];

int fun(int d[],int s[],int n,int i,int k){
    if(i==n) return 0;
    if(dp[i][k]!=-1) return dp[i][k];
    int ans = LLONG_MAX;
    repA(j,0,k){
        if(i+j>=n) break;
        int cost = s[i]*(d[i+j+1]-d[i]);
        ans = min(ans,cost+fun(d,s,n,i+j+1,k-j));
    }
    return dp[i][k] = ans;
}

void pre(){

}

void solve() {
    int n=0,l=0,k=0;
    cin>>n>>l>>k;
    int d[n+1],s[n];
    rep(i,n) cin>>d[i];
    rep(i,n) cin>>s[i];
    d[n] = l;
    memset(dp,-1,sizeof(dp));
    cout<<fun(d,s,n,0,k);
}

int32_t main(){
    fast
    int t=1;
//    cin>>t;
    pre();
    rep(i,t){
//        cout<<"case #"<<i+1<<": ";
        solve();
    }
    return 0;
}