Submission #397077

#	Time	Username	Problem	Language	Result	Execution time	Memory
397077		SavicS	Triple Jump (JOI19_jumps)	C++14	100 / 100	1638 ms	111712 KiB

jumps

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

#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)

using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn = 500005;
const int inf = 1e9 + 5;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

int n, q;
ll niz[maxn];

int L[maxn];
int R[maxn];

vector<pii> upit[maxn];
vector<pii> ja[maxn];

int res[maxn];

ll bor[4 * maxn][2]; // 0 - vrednost, 1 - vrednost + koliko_sam_dodao
ll lenj[4 * maxn];
void build(int v, int tl, int tr){
	bor[v][1] = -inf;
	if(tl == tr){
		bor[v][0] = niz[tl];
		return;
	}
	int mid = (tl + tr) / 2;
	build(v * 2, tl, mid);
	build(v * 2 + 1, mid + 1, tr);
	bor[v][0] = max(bor[v * 2][0], bor[v * 2 + 1][0]);
}

void propagate(int v, int tl, int tr){
	if(lenj[v]){
		bor[v][1] = max(bor[v][1], bor[v][0] + lenj[v]);
		if(tl != tr){
			lenj[v * 2] = max(lenj[v * 2], lenj[v]);
			lenj[v * 2 + 1] = max(lenj[v * 2 + 1], lenj[v]); 
		}
		lenj[v] = 0;
	}
}

void lazyupd(int v, int tl, int tr, int l, int r, ll val){
	propagate(v, tl, tr);
	if(tl > tr || l > tr || tl > r)return;
	if(tl >= l && tr <= r){
		bor[v][1] = max(bor[v][1], bor[v][0] + val);
		if(tl != tr){
			lenj[v * 2] = max(lenj[v * 2], val);
			lenj[v * 2 + 1] = max(lenj[v * 2 + 1], val);
		}
		return;
	}
	int mid = (tl + tr) / 2;
	lazyupd(v * 2, tl, mid, l, r, val);
	lazyupd(v * 2 + 1, mid + 1, tr, l, r, val);
	bor[v][1] = max(bor[v * 2][1], bor[v * 2 + 1][1]);
}

ll kveri(int v, int tl, int tr, int l, int r){
	propagate(v, tl, tr);
	if(l > tr || tl > r)return 0;
	if(tl >= l && tr <= r)return bor[v][1];
	int mid = (tl + tr) / 2;
	return max(kveri(v * 2, tl, mid, l, r), kveri(v * 2 + 1, mid + 1, tr, l, r));
}


void add(int X, int Y){
	if(X != 0 && Y != 0 && 2 * Y - X <= n){
		ja[X].pb({2 * Y - X, niz[X] + niz[Y]});
	}
}

int main()
{
   	ios::sync_with_stdio(false);
   	cout.tie(nullptr);
  	cin.tie(nullptr);
	cin >> n;
	ff(i,1,n)cin >> niz[i];
	
	stack<int> stek;
	fb(i,n,1){
		while(!stek.empty() && niz[stek.top()] <= niz[i]){
			L[stek.top()] = i;
			stek.pop();
		}
		stek.push(i);		
	}
	
	while(!stek.empty()){
		stek.pop();
	}
	
	ff(i,1,n){
		while(!stek.empty() && niz[stek.top()] <= niz[i]){
			R[stek.top()] = i;
			stek.pop();
		}
		stek.push(i);
	}
	
	// (i, i + 1), (i, R[i]), (L[i], i) - to su parovi za A i B
			
	ff(i,1,n){
		add(i, i + 1);
		add(i, R[i]);
		add(L[i], i);
	}
	
	cin >> q;
	ff(i,1,q){
		int l, r;
		cin >> l >> r;
		upit[l].pb({r, i});
	}
	
	build(1,1,n);
	fb(i,n,1){
		for(auto c : ja[i]){
			lazyupd(1,1,n,c.fi,n,c.se);
		}
		
		// ako hocu da i bude C, za i vazi da mogu da uzmem bilo koju vrednost levo
		
		for(auto c : upit[i]){
			int r = c.fi;
			int id = c.se;
			res[id] = kveri(1,1,n,i,r);
		}
		
	}	
	
	ff(i,1,q)cout << res[i] << '\n';
		
   	return 0;
}
/**

5
5 2 1 5 3
3
1 4
2 5
1 5

// probati bojenje sahovski ili slicno

**/

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