제출 #569773

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
569773		Lobo	밀림 점프 (APIO21_jumps)	C++17	0 / 100	190 ms	156116 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];

pair<int,int> tr[8*maxn];
int f1[8*maxn], f2[8*maxn], bg[maxn], cnt = 0;
int att(int no, int l, int r, int no1, int pos, int val) {
    if(no == 0) no = ++cnt;
    if(l > pos || r < pos) return no;
    if(l == r) {
        tr[no] = mp(val,l);
        return no;
    }

    int mid = (l+r)/2;
    if(pos <= mid) {
        //vai para esquerda
        f1[no] = att(f1[no],l,mid,f1[no1],pos,val);
        f2[no] = f2[no1];
    }
    else {

        f2[no] = att(f2[no],mid+1,r,f2[no1],pos,val);
        f1[no] = f1[no1];
    }
    tr[no] = max(tr[f1[no]],tr[f2[no]]);
    return no;
}

pair<int,int> qrrmx(int no, int l, int r, int L, int R) {
    if(l > R || r < L) return mp(0,0);
    if(l >= L && r <= R) return tr[no];
    int mid = (l+r)/2;
    return max(qrrmx(f1[no],l,mid,L,R),qrrmx(f2[no],mid+1,r,L,R));
}

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

    vector<pair<int,int>> vec;
    for(int i = 1; i <= n; i++) {
        h[i] = H[i-1];
        vec.pb(mp(h[i],i));
    }
    sort(all(vec));
    int ant = 0;
    for(auto X : vec) {
        ant = att(0,1,n,ant,X.sc,X.fr);
        bg[X.fr] = ant;
    }

    //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

    //pego o maior cara entre a e b < c
    //na vez do h[c] qual era o maior do intervalo
    
    // int s = qrrmx(bg[h[C]],1,n,A,B).sc;
    int s = 0;
    int mx = 0;
    for(int i = A; i <= B; i++) {
        if(h[i] < h[C] && h[i] > mx) {
            mx = h[i];
            s = i;
        }
    }
    if(s == 0) return -1;
    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: 0 / 23
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• Subtask 7: 0 / 19