Submission #569757

#	Time	Username	Problem	Language	Result	Execution time	Memory
569757		Lobo	Rainforest Jumps (APIO21_jumps)	C++17	23 / 100	1116 ms	63064 KiB

jumps

#include "jumps.h"
#include<bits/stdc++.h>
using namespace std;

const long long inf = (long long) 1e18 + 10;
const int inf1 = (int) 1e9 + 10;
#define dbl long double
#define endl '\n'
#define sc second
#define fr first
#define mp make_pair
#define pb push_back
#define all(x) x.begin(), x.end()

const int maxn = 2e5+10;

int n, h[maxn], nx[maxn][25], nxl[maxn][25], nxr[maxn][25];


void init(int N, vector<int> H) {
    n = N;

    for(int i = 1; i <= n; i++) {
        h[i] = H[i-1];
    }

    //os proximos da esquerda e direita de cada um e 0 se não tiver
    stack<pair<int,int>> stl;
    for(int i = 1; i <= n; i++) {
        while(stl.size() && stl.top().fr < h[i]) {
            stl.pop();
        }

        if(stl.size()) nxl[i][0] = stl.top().sc;
        else nxl[i][0] = 0;

        stl.push(mp(h[i],i));
    }
    stack<pair<int,int>> str;
    for(int i = n; i >= 1; i--) {
        while(str.size() && str.top().fr < h[i]) {
            str.pop();
        }

        if(str.size()) nxr[i][0] = str.top().sc;
        else nxr[i][0] = 0;

        str.push(mp(h[i],i));
    }

    //sei qual o proximo normal de cada um
    for(int i = 1; i <= n; i++) {
        if(nxr[i][0] == 0 || h[nxl[i][0]] > h[nxr[i][0]]) nx[i][0] = nxl[i][0];
        else nx[i][0] = nxr[i][0];

        // cout << i << " " << nx[i][0] << " " << nxl[i][0] << " " << nxr[i][0] << endl;
    }

    for(int pt = 1; pt <= 20; pt++) {
        for(int i = 1; i <= n; i++) {
            nx[i][pt] = nx[nx[i][pt-1]][pt-1];
        }
    }
    for(int pt = 1; pt <= 20; pt++) {
        for(int i = 1; i <= n; i++) {
            nxl[i][pt] = nxl[nxl[i][pt-1]][pt-1];
        }
    }
    for(int pt = 1; pt <= 20; pt++) {
        for(int i = 1; i <= n; i++) {
            nxr[i][pt] = nxr[nxr[i][pt-1]][pt-1];
        }
    }
}

int minimum_jumps(int A, int B, int C, int D) {
    A++;
    B++;
    C++;
    D++;
    //ir do a para o c

    int s = A;
    int t = C;
    int ans = 0;
    for(int pt = 20; pt >= 0; pt--) {
        if(nx[s][pt] == 0 || h[nx[s][pt]] > h[t]) continue;
        ans+= (1<<pt);
        s = nx[s][pt];
    }

    //vai so para a direita ate chegar no t
    for(int pt = 20; pt >= 0; pt--) {
        if(nxr[s][pt] == 0 || nxr[s][pt] > t) continue;
        ans+= (1<<pt);
        s = nxr[s][pt];
    }

    if(s == t) return ans;
    else return -1;
}

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