Submission #1207881

#	Time	Username	Problem	Language	Result	Execution time	Memory
1207881		countless	Rainforest Jumps (APIO21_jumps)	C++20	100 / 100	576 ms	51808 KiB

jumps

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

typedef long long ll;
typedef long double ld;

const ll MOD = 998244353;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp <<" "<<
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

int n, logn = 20;
vector<int> h;
vector<vector<int>> l, r, b;

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

    h.insert(h.begin(), INT_MAX);
    h.insert(h.end(), INT_MAX);

    n += 2;

    l = r = b = vector<vector<int>>(logn, vector<int>(n));

    vector<int> st;
    REP(i, 0, n) {
        while (!st.empty() and h[st.back()] <= h[i]) {
            st.pop_back();
        }

        l[0][i] = st.empty() ? i : st.back();
        st.push_back(i);
    }

    st.clear();

    for (int i = n-1; i >= 0; i--) {
        while (!st.empty() and h[st.back()] <= h[i]) {
            st.pop_back();
        }

        r[0][i] = st.empty() ? i : st.back();
        st.push_back(i);
    }

    REP(i, 0, n) {
        b[0][i] = h[l[0][i]] > h[r[0][i]] ? l[0][i] : r[0][i];
    }

    REP(i, 1, logn) {
        REP(j, 0, n) {
            l[i][j] = l[i-1][ l[i-1][j] ];
            r[i][j] = r[i-1][ r[i-1][j] ];
            b[i][j] = b[i-1][ b[i-1][j] ];
        }
    }
}

int minimum_jumps(int A, int B, int C, int D) {
    ++A, ++B, ++C, ++D;

    int curr = B;

    // no middle
    if (B + 1 == C) {
        return r[0][curr] <= D ? 1 : -1;
    }

    // find middle
    int mid = B+1;
    for (int i = logn-1; i >= 0; i--) {
        if (r[i][mid] < C) {
            mid = r[i][mid];
        }
    }

    // cerr << "here" sp curr sp mid sp C sp D << endl;
    // irrelevant
    if (h[curr] > h[mid]) {
        return r[0][curr] <= D ? 1 : -1;
    }

    // go to up below mid
    for (int i = logn-1; i >= 0; i--) {
        if (A <= l[i][curr] and h[ l[i][curr] ] < h[mid]) {
            curr = l[i][curr];
        }
    }

    int jumps = 0;
    // skip mid
    if (A <= l[0][curr]) {
        if (r[0][ l[0][curr] ] <= D) {
            return 1;
        }
    }

    // get to mid
    else {
        for (int i = logn-1; i >= 0; i--) {
            if (h[ b[i][curr] ] <= h[mid]) {
                curr = b[i][curr];
                jumps += (1 << i);
            }
        }

        // assert(curr == mid);
        if (curr == mid) {
            return r[0][curr] <= D ? jumps + 1 : -1;
        }
        
        // there is a shorter path that doesn't go to mid
        else {
            if (l[0][curr] > 0 and r[0][ l[0][curr] ] <= D) {
                return jumps + 2;
            }
        }
    }

    // go to C, D
    for (int i = logn-1; i >= 0; i--) {
        if (r[i][curr] < C) {
            curr = r[i][curr];
            jumps += (1 << i);
        }
    }

    // cerr << "here" sp curr sp mid sp C sp D << endl;
    return C <= r[0][curr] and r[0][curr] <= D ? jumps + 1 : -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