Submission #1078287

#	Time	Username	Problem	Language	Result	Execution time	Memory
1078287		GrindMachine	Bitaro's travel (JOI23_travel)	C++17	5 / 100	3096 ms	8540 KiB

travel

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

read the solution a long time ago, remember the key ideas

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n; cin >> n;
    vector<ll> a(n);
    rep(i,n) cin >> a[i];
    
    vector<ll> val1(n);
    rep(i,n-1) val1[i] = 2*a[i+1]-a[i];

    auto rmq = [&](ll l, ll r){
        ll mx = 0;
        for(int i = l; i <= r; ++i){
            amax(mx,val1[i]);
        }
        return mx;
    };

    vector<ll> val2(n);
    rep1(i,n-1) val2[i] = a[i]-2*a[i-1];

    vector<ll> b(n);
    rep(i,n){
        ll l = i, r = i, pos = i;
        ll cost = 0;

        while(!(l == 0 and r == n-1)){
            ll left_cost = inf2, right_cost = inf2;
            if(l-1 >= 0){
                left_cost = a[pos]-a[l-1];
            }
            if(r+1 < n){
                right_cost = a[r+1]-a[pos];
            }

            if(left_cost <= right_cost){
                cost += left_cost;
                ll lo = 0, hi = l-2;
                ll best = l-1;
                ll x = inf2;
                if(r+1 < n) x = a[r+1];
            
                while(lo <= hi){
                    ll mid = (lo+hi) >> 1;
                    if(rmq(mid,l-2) <= x){
                        best = mid;
                        hi = mid-1;
                    }
                    else{
                        lo = mid+1;
                    }
                }

                cost += a[l-1]-a[best];
                pos = best;
                l = best;
            }
            else{
                cost += right_cost;
                pos = r+1;
                r++;
            }
        }

        b[i] = cost;
    }

    ll q; cin >> q;
    while(q--){
        ll x; cin >> x;
        ll p = lower_bound(all(a),x)-a.begin();
        ll right_cost = inf2, left_cost = inf2;
        if(p < n){
            right_cost = abs(a[p]-x);
        }
        if(p-1 >= 0){
            left_cost = abs(a[p-1]-x);
        }

        ll ans = min(left_cost,right_cost);

        if(left_cost <= right_cost){
            ans += b[p-1];
        }
        else{
            ans += b[p];
        }

        cout << ans << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 30
• Subtask 4: 0 / 55