Submission #834922

#	Time	Username	Problem	Language	Result	Execution time	Memory
834922		maomao90	Triple Jump (JOI19_jumps)	C++17	32 / 100	4046 ms	4724 KiB

jumps

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

#define REP(i, s, e) for (int i = (s); i < (e); i++)
#define RREP(i, s, e) for (int i = (s); i >= (e); i--)
template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}

typedef unsigned long long ull;
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define SZ(_a) (int) _a.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 500005;

int n;
int a[MAXN];
int q;
int mx[MAXN];

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n;
    REP (i, 1, n + 1) {
        cin >> a[i];
    }
    cin >> q;
    REP (i, 1, q + 1) {
        int l, r; cin >> l >> r;
        REP (j, l, r + 1) {
            mx[j] = -INF;
        }
        vi stk;
        REP (j, l, r + 1) {
            while (!stk.empty() && a[stk.back()] < a[j]) {
                // c - j = j - stk.back()
                // c = 2 * j - stk.back()
                mxto(mx[2 * j - stk.back()], a[j] + a[stk.back()]);
                stk.pop_back();
            }
            if (!stk.empty()) {
                mxto(mx[2 * j - stk.back()], a[j] + a[stk.back()]);
            }
            stk.pb(j);
        }
        stk.clear();
        RREP (j, r, l) {
            while (!stk.empty() && a[stk.back()] < a[j]) {
                // c - stk.back() = stk.back() - j
                // c = 2 * stk.back() - j
                mxto(mx[2 * stk.back() - j], a[j] + a[stk.back()]);
                stk.pop_back();
            }
            if (!stk.empty()) {
                mxto(mx[2 * stk.back() - j], a[j] + a[stk.back()]);
            }
            stk.pb(j);
        }
        ll ans = 0;
        REP (j, l + 1, r + 1) {
            mxto(mx[j], mx[j - 1]);
            mxto(ans, (ll) a[j] + mx[j]);
        }
        cout << ans << '\n';
    }
    return 0;
}

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