Submission #746472

#	Time	Username	Problem	Language	Result	Execution time	Memory
746472		GrindMachine	Triple Jump (JOI19_jumps)	C++17	27 / 100	37 ms	11976 KiB

jumps

// Om Namah Shivaya

#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) 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(x) 42
#endif

/*

refs:
https://codeforces.com/blog/entry/68269?#comment-527139
edi


choose l <= a < b < c <= r s.t:
b-a <= c-b
arr[a]+arr[b]+arr[c] is max

let's say we fix (a,b)
when can we consider (a,b) as an option
if there is a guy in between with a higher val, then we dont have to consider pair (a,b)

if there exists k s.t a < k < b with arr[k] >= arr[a] or arr[k] >= arr[b], then we can do this:
arr[k] >= arr[a]: set a = k
arr[k] >= arr[b]: set b = k

by doing this, we reduce the jump distance from a to b
so we have more options for c

in short, consider pair (a,b) only if there is nobody in between with a higher or equal value

how many such pairs are there?

iterate over b from 1 to n and find all valid a

maintain all valid a values when increasing b
once some a becomes bad (some >= guy appears in between), he never becomes good for bigger b

we want to find all a for whom nobody >= them has appeared in [a,b] so far

when moving to a new b, we can make pairs (a,b) for all active a
then we will remove all a values for which arr[a] <= arr[b]

(yet to complete explanation)

*/

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> arr(n+5);
    rep1(i,n) cin >> arr[i];

    vector<ll> good;
    vector<pll> pairs;

    rep1(b,n){
        while(!good.empty() and arr[b] >= arr[good.back()]){
            pairs.pb({good.back(), b});
            good.pop_back();
        }

        if(!good.empty()){
            pairs.pb({good.back(), b});
        }
    
        good.pb(b);
    }

    vector<ll> suff(n+5);
    rev(i,n,1) suff[i] = max(suff[i+1], arr[i]);

    ll q; cin >> q;

    while(q--){
        ll ans = 0;
        for(auto [a, b] : pairs){
            ll dis = b - a;
            ll min_c = b + dis;
            if(min_c <= n){
                ll val = arr[a] + arr[b] + suff[min_c];
                amax(ans, val);
            }
        }

        cout << ans << endl;
    }
}

int main()
{
    fastio;

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

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

    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 14
• Subtask 3: 27 / 27
• Subtask 4: 0 / 54