Submission #675993

#	Time	Username	Problem	Language	Result	Execution time	Memory
675993		ghostwriter	Triple Jump (JOI19_jumps)	C++17	100 / 100	1107 ms	104492 KiB

jumps

// #pragma GCC optimize ("Ofast")
// #pragma GCC target ("avx2")
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
    #include <debug.h>
#else
    #define debug(...)
#endif
#define ft front
#define bk back
#define st first
#define nd second
#define ins insert
#define ers erase
#define pb push_back
#define pf push_front
#define _pb pop_back
#define _pf pop_front
#define lb lower_bound
#define ub upper_bound
#define mtp make_tuple
#define bg begin
#define ed end
#define all(x) (x).bg(), (x).ed()
#define sz(x) (int)(x).size()
// #define int long long
typedef long long ll; typedef unsigned long long ull;
typedef double db; typedef long double ldb;
typedef pair<int, int> pi; typedef pair<ll, ll> pll;
typedef vector<int> vi; typedef vector<ll> vll; typedef vector<pi> vpi; typedef vector<pll> vpll;
typedef string str;
#define FOR(i, l, r) for (int i = (l); i <= (r); ++i)
#define FOS(i, r, l) for (int i = (r); i >= (l); --i)
#define FRN(i, n) for (int i = 0; i < (n); ++i)
#define FSN(i, n) for (int i = (n) - 1; i >= 0; --i)
#define EACH(i, x) for (auto &i : (x))
#define WHILE while
template<typename T> T gcd(T a, T b) { T d2 = (a | b) & -(a | b); a /= d2; b /= d2; WHILE(b) { a = a % b; swap(a, b); } return a * d2; }
template<typename T> T lcm(T a, T b) { return a / gcd(a, b) * b; }
void _assert(bool statement) { if (statement) return; cerr << "\n>> Assertion failed!\n"; exit(0); }
void _assert(bool statement, const str &message) { if (statement) return; cerr << "\n>> Assertion failed: " << message << '\n'; exit(0); }
void _error(const str &message) { cerr << "\n>> Error: " << message << '\n'; exit(0); }
#define file "TEST"
mt19937 rd(chrono::steady_clock::now().time_since_epoch().count());
ll rand(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rd); }
/*
----------------------------------------------------------------
    END OF TEMPLATE
----------------------------------------------------------------
    Tran The Bao - ghostwriter
    Training for VOI23 gold medal
----------------------------------------------------------------
    GOAT
----------------------------------------------------------------
*/
const int N = 5e5 + 5;
int n, q, a[N];
pi query[N];
void input(int test_id) {
	cin >> n;
	FOR(i, 1, n) cin >> a[i];
	cin >> q;
	FOR(i, 1, q) cin >> query[i].st >> query[i].nd;
}
namespace subtask12 {
	const int N = 5005;
	int d[N][N], maxx[N], pos[N], rs = 0;
	int get(int l, int r) {
		if (!pos[l]) {
			maxx[l] = a[l];
			pos[l] = l;
		}
		WHILE(pos[l] < r) {
			++pos[l];
			maxx[l] = max(maxx[l], a[pos[l]]);
		}
		return maxx[l];
	}
	void solve() {
		FOR(len, 3, n)
		FOR(l, 1, n - len + 1) {
			int r = l + len - 1;
			d[l][r] = max({d[l + 1][r], d[l][r - 1], a[l] + a[r] + get(l + 1, l + (r - l) / 2)});
		}
		FOR(i, 1, q) {
			int l = query[i].st, r = query[i].nd;
			cout << d[l][r] << '\n';
		}
	}
}
namespace subtask3 {
	int maxx[N][18], LOG[N], rs = 0;
	multiset<pi> s;
	void build() {
		FOR(i, 1, n) {
			maxx[i][0] = a[i];
			LOG[i] = log2(i);
		}
		FOR(j, 1, 17)
		FOR(i, 1, n) {
			if (i + (1 << j) - 1 > n) break;
			maxx[i][j] = max(maxx[i][j - 1], maxx[i + (1 << (j - 1))][j - 1]);
		}
	}
	int get(int l, int r) {
		int l2 = LOG[r - l + 1];
		return max(maxx[l][l2], maxx[r - (1 << l2) + 1][l2]);
	}
	int cal(int i) {
		int rs = 0;
		FOR(j, 1, i - 2) {
			int mid = j + (i - j) / 2;
			rs = max(rs, a[j] + a[i] + get(j + 1, mid));
		}
		FOR(j, i + 2, n) {
			int mid = i + (j - i) / 2;
			rs = max(rs, a[i] + a[j] + get(i + 1, mid));
		}
		FOR(j, 1, i - 1)
			if (2 * i - j <= n)
				rs = max(rs, a[j] + a[i] + get(2 * i - j, n));
		return rs;
	}
	void solve() {
		FOR(i, 1, n) {
			s.ins({a[i], i});
			if (sz(s) > 18) s.ers(s.bg());
		}
		build();
		EACH(j, s) {
			int i = j.nd;
			rs = max(rs, cal(i));
		}
		cout << rs;
	}
}
namespace subtask4 {
	/*
		This is a tough problem, although I have a beautiful observation for subtask3 but still fall to the trap.
		Consider i and j, a[i] and a[j] must be bigger than every element that lies in [i + 1, j - 1] (as our claim) or else, we will
		move i (or j) to the position that is bigger than a[i] (or a[j]) and the cost won't change.
		What happen when we consider all pair i, j that sastified the condition? The answer is at most 2 * n. For all i, first we look at i - 1,
		after that the first element bigger than a[i - 1] and continue... This is familar? A standard deque trick (monotonic stack trick).
		The number of candidate pair (i, j) is at most 2 * n, now how to solve the problem with this observation is much more of a problem!
		I've seen some submission that use some tricks to avoid lazy propagate but I think it's too complicated so I'll use lazy propagate segtree to
		update and answer queries in a offline way.
		Upd: Nah, that lazy avoiding trick isn't that hard, it's just because I'm so high right now :v.
	*/
	const int oo = 5e8 + 5;
	struct Node {
		int ab, c, ans;
		Node() : ab(-oo), c(-oo), ans(-oo) {}
		Node(int ab, int c, int ans) : ab(ab), c(c), ans(ans) {}
	};
	const int T = 2e6 + 5;
	int qn[N], lp[N], ans[N];
	vpi a1;
	Node tr[T];
	Node comb(const Node &a, const Node &b) { return Node(max(a.ab, b.ab), max(a.c, b.c), max({a.ans, b.ans, a.ab + b.c})); }
	void upd(int i, int l, int r, int q, int v, bool type) {
		if (r < q || l > q) return;
		if (l == r) {
			if (!type) tr[i].c = max(tr[i].c, v);
			else tr[i] = Node(max(tr[i].ab, v), tr[i].c, max(tr[i].ans, v + tr[i].c));
			return;
		}
		int mid = l + (r - l) / 2;
		if (q <= mid) upd(i * 2, l, mid, q, v, type);
		else upd(i * 2 + 1, mid + 1, r, q, v, type);
		tr[i] = comb(tr[i * 2], tr[i * 2 + 1]);
	}
	Node get(int i, int l, int r, int ql, int qr) {
		if (r < ql || l > qr) return Node();
		if (ql <= l && r <= qr) return tr[i];
		int mid = l + (r - l) / 2;
		return comb(get(i * 2, l, mid, ql, qr), get(i * 2 + 1, mid + 1, r, ql, qr));
	}
	void solve() {
		FOR(i, 1, n) {
			int cur = i - 1;
			WHILE(cur) {
				a1.pb({cur, i});
				if (a[cur] > a[i]) {
					lp[i] = cur;
					break;
				}
				cur = lp[cur];
			}
		}
		sort(all(a1), [&](const pi &a, const pi &b) -> bool { return a.st > b.st; });
		FOR(i, 1, q) qn[i] = i;
		sort(qn + 1, qn + 1 + q, [&](const int &a, const int &b) { return query[a].st > query[b].st; });
		int cp = 0;
		FOR(i, 1, n) upd(1, 1, n, i, a[i], 0);
		FOR(j, 1, q) {
			int i = qn[j], l = query[i].st, r = query[i].nd;
			WHILE(cp < sz(a1) && a1[cp].st >= l) {
				upd(1, 1, n, 2 * a1[cp].nd - a1[cp].st, a[a1[cp].st] + a[a1[cp].nd], 1);
				++cp;
			}
			ans[i] = get(1, 1, n, l, r).ans;
		}
		FOR(i, 1, q) cout << ans[i] << '\n';
	}
}
void solve(int test_id) {
	if (n <= 5000) {
		subtask12::solve();
		return;
	}
	if (n <= 2e5 && q == 1 && query[1].st == 1 && query[1].nd == n) {
		subtask3::solve();
		return;
	}
	subtask4::solve();
}
void reinit(int test_id) {

}
signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    // freopen(file".inp", "r", stdin);
    // freopen(file".out", "w", stdout);
    int test_num = 1;
    // cin >> test_num; // comment if the problem does not requires multitest
    FOR(i, 1, test_num) {
        input(i); // input in noninteractive problems for case #i
        solve(i); // main function to solve case #i
        reinit(i); // reinit global data to default used in case #i
    }
    #ifdef LOCAL
        cerr << "\nTime: " << setprecision(5) << fixed << (ldb)clock() / CLOCKS_PER_SEC << "ms.\n";
    #endif
    return 0;
}
/*
5
5 2 1 5 3
1
1 5
----------------------------------------------------------------
From Benq:
    stuff you should look for
        * int overflow, array bounds
        * special cases (n=1?)
        * do smth instead of nothing and stay organized
        * WRITE STUFF DOWN
        * DON'T GET STUCK ON ONE APPROACH
----------------------------------------------------------------
*/

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