제출 #1318769

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1318769		vaishakhv	Safety (NOI18_safety)	C++20	0 / 100	11 ms	1452 KiB

safety

// Source: https://usaco.guide/general/io

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define eb emplace_back // faster than push_back xD

// pbds UwU
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
#define oset tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update> // use pair for ms

// my io library :D
#define m1(x) template<class T, class... U> void x(T&& a, U&&... b)
#define m2(x) (ll[]){(x forward<U>(b),0)...}

m1(pr){cout << forward<T>(a);  m2(cout << " " <<); cout << "\n";}
m1(re){cin >> forward<T>(a); m2(cin >>);}

ll n, h, base = 1;
vector<ll> treeseg;

ll rangemin(ll a, ll b){
    a += base; b += base;
    ll s = 1e18;
    while (a <= b){
        if (a%2 == 1) s = min(s, treeseg[a++]);
        if (b%2 == 0) s = min(s, treeseg[b--]);

        a /= 2; b /= 2;
    }
    return s;
}

int main() {
	ios::sync_with_stdio(0);
    cin.tie(0);

    re(n, h);

    vector<ll> s(n);
    for (ll i{}; i < n; i++) re(s[i]);

    ll maxl = max({*max_element(s.begin(), s.end()),  s[0] + (n-1)*h, s[n-1] + (n-1)*h});
    while (base <= maxl) base *= 2;
    treeseg.resize(2*base);

    vector<vector<ll>> dp(n, vector<ll>(maxl+1, 1e18));

    for (ll i{}; i <= maxl; i++){
        dp[0][i] = abs(s[0]-i);
    }

    for (ll i = 1; i < n; i++){
        fill(treeseg.begin(), treeseg.end(), (ll)1e18);

        ll low = max(0LL, s[0] - i*h);
        ll high = s[0] + i*h;

        for (ll k = low; k <= high; k++){
            treeseg[base+k] = dp[i-1][k];
        }
        for (ll k = base-1; k > 0; k--){
            treeseg[k] = min(treeseg[2*k], treeseg[2*k+1]);
        }
        for (ll j = low; j <= high; j++){
            ll l = max(low, j-h), r = min(high, j+h);
            dp[i][j] = rangemin(l, r) + abs(j-s[i]);
        }
    }

    ll low = max(0LL, s[0] - (n-1)*h);
    ll high = s[0] + (n-1)*h;

    ll ans = 1e18;
    for (ll i = low; i <= high; i++) ans = min(ans, dp[n-1][i]);
    pr(ans);
}

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• Subtask 3: 0 / 9
• Subtask 4: 0 / 5
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