제출 #555818

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
555818		joelau	밀림 점프 (APIO21_jumps)	C++14	48 / 100	1254 ms	70912 KiB

jumps

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

int n, p1[200005][20], p2[200005][20], p3[200005][20], p4[200005][20], X[200005];
vector<int> lst[200005];
stack<int> s;

void init(int N, std::vector<int> H) {
    n = N;
    for (int i = 0; i < N; ++i) X[i] = H[i];
    memset(p1,-1,sizeof(p1));
    memset(p2,-1,sizeof(p2));
    memset(p3,-1,sizeof(p3));
    memset(p4,-1,sizeof(p4));
    for (int i = 0; i < N; ++i) {
        while (!s.empty() && H[i] >= H[s.top()]) s.pop();
        if (!s.empty()) p1[i][0] = p3[i][0] = s.top();
        s.push(i);
    }
    while (!s.empty()) s.pop();
    for (int i = N-1; i >= 0; --i) {
        while (!s.empty() && H[i] >= H[s.top()]) s.pop();
        if (!s.empty()) p2[i][0] = p4[i][0] = s.top();
        s.push(i);
    }
    for (int i = 0; i < N; ++i) if (p1[i][0] == -1 || (p1[i][0] != -1 && p2[i][0] != -1 && H[p1[i][0]] < H[p2[i][0]])) swap(p1[i][0], p2[i][0]);
    for (int k = 1; k < 20; ++k) for (int i = 0; i < N; ++i) {
        if (p1[i][k-1] == -1) p1[i][k] = -1;
        else p1[i][k] = p1[p1[i][k-1]][k-1];
        if (p2[i][k-1] == -1) p2[i][k] = -1;
        else p2[i][k] = p2[p2[i][k-1]][k-1];
        if (p3[i][k-1] == -1) p3[i][k] = -1;
        else p3[i][k] = p3[p3[i][k-1]][k-1];
        if (p4[i][k-1] == -1) p4[i][k] = -1;
        else p4[i][k] = p4[p4[i][k-1]][k-1];
    }
}

int minimum_jumps(int A, int B, int C, int D) {
    int x = C;
    for (int k = 19; k >= 0; --k) if (p3[x][k] != -1 && p3[x][k] > B) x = p3[x][k];
    x = p3[x][0];
    if (x >= A && x <= B && p4[x][0] >= C && p4[x][0] <= D) return 1;
    int y = B;
    for (int k = 19; k >= 0; --k) if (p4[y][k] != -1 && p4[y][k] < C) y = p4[y][k];
    y = p4[y][0];
    if (y < C || y > D) return -1;
    x = B;
    for (int k = 19; k >= 0; --k) if (p3[x][k] != -1 && X[p3[x][k]] <= X[y] && p3[x][k] >= A) x = p3[x][k];
    int ans = 0;
    for (int k = 19; k >= 0; --k) if (p1[x][k] != -1 && X[p1[x][k]] <= X[y]) ans += (1<<k), x = p1[x][k];
    for (int k = 19; k >= 0; --k) if (p2[x][k] != -1 && X[p2[x][k]] <= X[y]) ans += (1<<k), x = p2[x][k];
    if (x != y) return -1;
    else return ans;
}

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