제출 #814668

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
814668		sentheta	밀림 점프 (APIO21_jumps)	C++17	4 / 100	1278 ms	114660 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], lo[N][LN];


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

	
}

int minimum_jumps(int A,int B,int C,int D){
	
	// 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] that can reach [C..D]
	int s = B;
	for(int b=LN-1; b>=0; b--){
		int x = pl[s][b];
		if(A<=x && a[x]<=a[t]) s = x;
	}
	// impossible
	if(!(a[s] < a[t])){
		return -1;
	}
	// directly connected
	for(int x : adj[s]){
		if(C<=x && x<=D) return 1;
	}

	
	int ans = 0;
	
	// find closest ancestor that is directly connected
	
	// hi edges
	for(int b=LN-1; b>=0; b--){
		bool ok = 1;
		int x = hi[s][b];
		
		ok &= a[x]<a[t];
		ok &= !(C<=x && x<=D);
		for(int y : adj[x]){
			ok &= !(C<=y && y<=D);
		}
		
		if(ok){
			s = hi[s][b];
			ans += 1<<b;
		}
	}
	
	// rightward edges
	// leftward is not possible, since it was skippable
	for(int b=LN-1; b>=0; b--){
		bool ok = 1;
		int x = pr[s][b];
		
		ok &= a[x]<a[t];
		ok &= !(C<=x && x<=D);
		for(int y : adj[x]){
			ok &= !(C<=y && y<=D);
		}
		
		if(ok){
			s = pr[s][b];
			ans += 1<<b;
		}
	}
	s = pr[s][0];
	ans += 2;
	
	assert(C<=pr[s][0] && pr[s][0]<=D);
	
		
	
	
	
	

	
	return ans;
}

• Subtask 1: 4 / 4
• Subtask 2: 0 / 8
• Subtask 3: 0 / 13
• Subtask 4: 0 / 12
• Subtask 5: 0 / 23
• Subtask 6: 0 / 21
• Subtask 7: 0 / 19