Cacher is the code snippet organizer for pro developers

#include<bits/stdc++.h>
using namespace std;

// #DynamicProgramming
// https://www.hackerearth.com/problem/algorithm/journey-1-0418ba3e/

int minCost(int n, vector<int> li, vector<int> ci, vector<int> pi, int x);
int solve(vector<int> li, vector<int> ci, vector<int> pi, int x, int start, int end, int cost);

int main(){
    int n,m;
    cin>>n>>m;
    vector<int> li(n-1);
    vector<int> ci(n-1);
    vector<int> pi(n-1);
    for(int i=0;i<n-1;i++){
        cin>>li[i];
    }
    for(int i=0;i<n-1;i++){
        cin>>ci[i];
    }
    for(int i=0;i<n-1;i++){
        cin>>pi[i];
    }
    while(m--){
        int x;
        cin>>x;
        cout<<minCost(n,li,ci,pi,x)<<endl;
    }
    return 0;
}

int solve(vector<int> li, vector<int> ci, vector<int> pi, int charge, int start, int end, int cost){
    if(charge<0){
        return -1;  // if charge is depleted even before reaching the city
    }
    if(charge<li[start]){
        if(ci[start]<li[start]){   // if both charge & battery at shop are not enough to go to next city
            return -1;
        }
    }
    if(start==(end-1)){ // at last-but-one city
        if(charge<li[start]){   // if current charge is less than that is req to travel to last city
            //charge=ci[start]-li[start]; // replace battery
            return cost+pi[start];  // buy the battery and move forward
        }else{
            //charge-=li[start];
            return cost;    // journey possible without buying battery
        }
    }
    int with,without;
    without=solve(li,ci,pi,charge-li[start],start+1,end,cost);
    with=solve(li,ci,pi,ci[start]-li[start],start+1,end,cost+pi[start]);
    if(without!=-1){   // if journey can be completed without buying battery at current city
        return min(with,without);
    }else{
        if(with!=-1){
            return with;    // if journey can be completed only if battery is bought at current city
        }else{
            return -1;    // if journey can't be completed even after buying
        }
    }
}
int minCost(int n,vector<int> li,vector<int> ci,vector<int> pi,int x){
    return solve(li,ci,pi,x,0,n-1,0);
}