제출 #814783

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
814783		sentheta	밀림 점프 (APIO21_jumps)	C++17	100 / 100	1187 ms	50532 KiB

jumps

#include "jumps.h"
#include<bits/stdc++.h>
#ifdef VANWIJ
	#define DBG 1
#else
	#define DBG 0
#endif
using namespace std;

#define Int long long
#define V vector
#define pii pair<int,int>
#define ff first
#define ss second

static mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

#define pow2(x) (1LL<<(x))
#define msb(x) (63-__builtin_clzll(x))
#define bitcnt(x) (__builtin_popcountll(x))

#define nl '\n'
#define _ << ' ' <<
#define all(x) (x).begin(), (x).end()
#define rep(i,a,b) for(int i = (int)(a); i < (int)(b); i++)

#define cerr if(DBG) cout
#define dbg(x) {cerr << "?" << #x << " : " << (x) << endl << flush;}

const int N = 2e5+5;
const int LN = 18;

int n;
int a[N];

V<int> adj[N];
int pl[N][LN], pr[N][LN];
int hi[N][LN];


void init(int _n, V<int> _a){
	n = _n;
	a[0] = a[n+1] = n+1;
	rep(i,1,n+1) a[i] = _a[i-1];
	
	// leftward edges
	V<int> stak;
	for(int i=0; i<=n+1; i++){
		while(!stak.empty() && a[stak.back()]<=a[i]) stak.pop_back();
		int j = stak.empty() ? i : stak.back();
		pl[i][0] = j;
		stak.push_back(i);
	}

	// rightward edges
	stak.clear();
	for(int i=n+1; i>=0; i--){
		while(!stak.empty() && a[stak.back()]<=a[i]) stak.pop_back();
		int j = stak.empty() ? i : stak.back();
		pr[i][0] = j;
		dbg(i); dbg(j); cerr << nl;
		stak.push_back(i);
	}
	
	for(int i=0; i<=n+1; i++){
		if(a[pl[i][0]] > a[pr[i][0]]){
			hi[i][0] = pl[i][0];
		}
		else{
			hi[i][0] = pr[i][0];
		}
	}
	
	dbg(pr[4][0]);
	dbg(pr[n+1][0]);
	
	rep(lg,1,LN)
	for(int i=0; i<=n+1; i++){
		pl[i][lg] = pl[pl[i][lg-1]][lg-1];
		pr[i][lg] = pr[pr[i][lg-1]][lg-1];
		hi[i][lg] = hi[hi[i][lg-1]][lg-1];
	}
}

int minimum_jumps(int A,int B,int C,int D){
	A++; B++; C++; D++;
	
	// touching
	if(B==C-1){
		if(pr[B][0] <= D) return 1;
		return -1;
	}
	
	// maximum in [B+1..C-1]
	int m = B+1;
	for(int b=LN-1; b>=0; b--){
		int x = pr[m][b];
		if(x<C) m = x;
	}
	if(a[B] > a[m]){
		if(pr[B][0] <= D) return 1;
		return -1;
	}
	
	
	// // maximum in [C..D]
	// int t = C;
	// for(int b=LN-1; b>=0; b--){
	// 	int x = pr[t][b];
	// 	if(x<=D) t = x;
	// }
	
	// maximum in [A..B] not over a[m]
	int s = B;
	for(int b=LN-1; b>=0; b--){
		int x = pl[s][b];
		if(A<=x && a[x]<a[m]) s = x;
	}
	
	// // impossible
	// if(!(a[s] < a[t])) return -1;
	// if(!(a[m] < a[t])) return -1;
	// // if jump over m
	// for(int x : adj[s]){
	// 	if(C<=x && x<=D) return 1;
	// }
	// WLOG a[s] < a[m]
	
	// jump over m
	
	int ans = 0;
	
	// there is better in interval
	if(A <= pl[s][0]){
		// if jump over
		int x = pr[pl[s][0]][0];
		if(C<=x && x<=D) return 1;
		// no need to go left since it will land at over a[t]
	}
	else{
		// hi edges, not passing a[m]
		for(int b=LN-1; b>=0; b--){
			int x = hi[s][b];
			if(a[x]<=a[m]){
				s = x;
				ans += 1<<b;
			}
		}
		
		// stopped at m
		if(s==m){
			if(C<=pr[s][0] && pr[s][0]<=D) return ans+1;
			return -1;
		}
		// if jump over
		{
			int x = pr[pl[s][0]][0];
			if(C<=x && x<=D) return ans+2;
		}
	}
	
	// rightward edges
	// leftward is not possible, since it coulve been skipped
	for(int b=LN-1; b>=0; b--){
		int x = pr[s][b];
		if(x < C){
			s = x;
			ans += 1<<b;
		}
	}
	s = pr[s][0];
	ans++;
	
	if(C<=s && s<=D) return ans;
	return -1;
}

• Subtask 1: 4 / 4
• Subtask 2: 8 / 8
• Subtask 3: 13 / 13
• Subtask 4: 12 / 12
• Subtask 5: 23 / 23
• Subtask 6: 21 / 21
• Subtask 7: 19 / 19