제출 #860410

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
860410		mychecksedad	밀림 점프 (APIO21_jumps)	C++17	100 / 100	851 ms	108548 KiB

jumps

#include<jumps.h>
#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD (1000000000+7)
#define MOD1 (998244353)
#define pb push_back
#define all(x) x.begin(), x.end()
#define en cout << '\n'
const int N = 1e6+100, M = 1e5+10, K = 22;


int n, L[N], R[N], up[N][K][2], dp[N][K], le[N][K], ri[N][K];
vector<int> H;
int maxx(int l, int r){
	int k = int(log2(r-l+1));
	return max(dp[l][k], dp[r-(1<<k)+1][k]);
}

void init(int _n, vector<int> h){
	n = _n;
	H = h;
	
	H.pb(MOD);

	deque<int> q;
	L[0] = n;
	q.pb(0);
	for(int i = 1; i < n; ++i){
		while(!q.empty() && H[q.back()] < H[i]) q.pop_back();
		if(!q.empty()) L[i] = q.back();
		else L[i] = n;
		q.pb(i);
	}
	
	q.clear();
	R[n-1] = n;
	q.pb(n-1);
	for(int i = n-2; i >= 0; --i){
		while(!q.empty() && H[q.back()] < H[i]) q.pop_back();
		if(!q.empty()) R[i] = q.back();
		else R[i] = n;
		q.pb(i);
	}

	for(int i = 0; i < n; ++i) dp[i+1][0] = H[i];
	for(int j = 1; j < K; ++j){
		for(int i = 1; i + (1<<j) <= n + 1; ++i){
			dp[i][j] = max(dp[i][j - 1], dp[i+(1<<(j-1))][j-1]);
		}
	}
	

	up[n][0][0] = n;
	up[n][0][1] = n;
	for(int i = 0; i < n; ++i){
		if(H[L[i]] > H[R[i]]){
			up[i][0][0] = L[i];
			up[i][0][1] = max(L[i], R[i]);
		}else{
			up[i][0][1] = max(L[i], R[i]);
			up[i][0][0] = R[i];
		}
		le[i][0] = L[i];
		ri[i][0] = R[i];
	}
	le[n][0] = n;
	ri[n][0] = n;
	for(int j = 1; j < K; ++j){
		for(int i = 0; i <= n; ++i){
			up[i][j][0] = up[up[i][j - 1][0]][j - 1][0];
			up[i][j][1] = max(up[i][j - 1][1], up[up[i][j - 1][0]][j - 1][1]);
			le[i][j] = le[le[i][j - 1]][j - 1];
			ri[i][j] = ri[ri[i][j - 1]][j - 1];
		}
	}

}
int minimum_jumps(int A, int B, int C, int D){
	int v = -1, ans = 0;
	int mx = maxx(C+1, D+1);
	
	v = B;
	for(int j = K - 1; j >= 0; --j){
		if(le[v][j] >= A && H[le[v][j]] < mx){
			v = le[v][j];
		}
	}
	if(H[v] > mx) return -1;

	for(int j = K - 1; j >= 0; --j){
		if(up[v][j][1] < C && H[up[v][j][0]] < mx){
			ans += (1<<j);
			v = up[v][j][0];
		}
	}
	for(int j = K - 1; j >= 0; --j){
		if(ri[v][j] < C){
			ans += (1<<j);
			v = ri[v][j];
		}
	}
	if(ri[v][0] >= C && ri[v][0] <= D){
		return ans + 1;
	}
	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