제출 #417433

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
417433		CSQ31	밀림 점프 (APIO21_jumps)	C++14	100 / 100	1223 ms	51340 KiB

jumps

#include "jumps.h"
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define sz(a) (int)(a.size())
typedef long long int ll;
typedef pair<int,int> pii;
typedef vector<vector<int>> vii;
const int MAXN = 2e5+5;
int l[20][MAXN],r[20][MAXN],high[20][MAXN];
vector<int>H;
void init(int n, vector<int>h){
	H = h;
	H.insert(H.begin(),1e9);
	H.pb(1e9);
	stack<pii>stk;
	for(int i=0;i<=n+1;i++){
		while(!stk.empty() && stk.top().fi <= H[i])stk.pop();
		if(!stk.empty())l[0][i] = stk.top().se;
		else l[0][i] = i;
		stk.push({H[i],i});
	}
	while(!stk.empty())stk.pop();
	for(int i=n+1;i>=0;i--){
	    while(!stk.empty() && stk.top().fi <= H[i])stk.pop();
		if(!stk.empty())r[0][i] = stk.top().se;
		else r[0][i] = i;
		stk.push({H[i],i});
	}
	for(int i=0;i<=n+1;i++){
		high[0][i] = (H[r[0][i]] > H[l[0][i]])?r[0][i]:l[0][i];
	}
	//cout<<high[0][1]<<'\n';
	for(int j=1;j<20;j++){
		for(int i=0;i<=n+1;i++){
			l[j][i] = l[j-1][l[j-1][i]];
			r[j][i] = r[j-1][r[j-1][i]];
			high[j][i] = high[j-1][high[j-1][i]];
		}
	}
	
}
//
int minimum_jumps(int a,int b,int c,int d){
	a++;
	b++;
	c++;
	d++;
	if(c-1 == b){
		return r[0][b] <=d?1:-1;
	}
	int mid = b+1; //mid just means bigger element in (b,c),cuz we have to cross mid regardless
	for(int i=19;i>=0;i--)if(r[i][mid] < c)mid = r[i][mid];
	if(r[0][mid] > d)return -1;
	if(H[b] > H[mid]){
		return r[0][b] <= d?1:-1;
	}
	int cur = b;
	for(int i=19;i>=0;i--)if(l[i][cur] >= a && H[l[i][cur]] < H[mid])cur = l[i][cur]; //cur is where we should start from
	int ans= 0;
	//if(a == 1)cout<<l[0][cur]<<'\n';
	if(l[0][cur] >= a){
		if(r[0][l[0][cur]] <= d)return 1; //if this case fails it just means we can never ever again go left
	}else{
		for(int i=19;i>=0;i--){
			if(H[high[i][cur]] <= H[mid]){
				cur = high[i][cur];
				ans+=(1<<i);
			}
		}
		//if(a == 1)cout<<cur<<" "<<ans<<'\n';
		if(cur == mid)return r[0][mid] <=d?ans+1:-1;
		else if(l[0][cur] && r[0][l[0][cur]] <= d)return ans+2; 
	}
	for(int i=19;i>=0;i--){
		if(r[i][cur] < c){
			cur = r[i][cur];
			ans+=(1<<i);
		}
	}
	cur = r[0][cur];
	if(c<=cur && cur<=d)return ans+1;
	else 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