제출 #497763

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
497763		Karliver	밀림 점프 (APIO21_jumps)	C++17	100 / 100	1131 ms	50460 KiB

jumps

    
#include <bits/stdc++.h>
//#include "jumps.h"
#define FIXED_FLOAT(x)  std::fixed <<std::setprecision(20) << (x)
#define all(v) (v).begin(), (v).end()
using namespace  std;
#define forn(i,n) for (int i = 0; i < (n); ++i)
#define rforn(i, n) for(int i = (n) - 1;i >= 0;--i)
#define sz(x) (int)x.size()
#define ff first
#define se second
#define mp make_pair
using ll = long long;
int mod = (ll)1e9 + 7;
const int INF = 1e9 + 1;

const double eps = 1e-7;

template <class T> using V = vector<T>;  
template <class T> using VV = V<V<T>>;  
template<class T, size_t SZ> using AR = array<T, SZ>;
template<class T> using PR = pair<T, T>;
template <typename XPAX>
bool ckma(XPAX &x, XPAX y) {
    return (x < y ? x = y, 1 : 0);
}
template <typename XPAX>
bool ckmi(XPAX &x, XPAX y) {
    return (x > y ? x = y, 1 : 0);
}

const int mxn = 2e5+2;


const int G = 20;
int F[mxn][G], L[mxn][G], R[mxn][G];

V<int> H;

int n;
void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)

void init(int N, V<int> a) {
    
    H = a;
    H.insert(H.begin(), INF);
    H.push_back(INF);
    //debug(H);
    n = N;
    stack<int> st;
    forn(i, N+2) {
        while(sz(st) && H[st.top()] <= H[i])
            st.pop();
        
        L[i][0] = sz(st) ? st.top() : i;
        st.push(i);
    }
    while(sz(st))st.pop();
    rforn(i, N+2) {
        while(sz(st) && H[st.top()] <= H[i])
            st.pop();
        
        R[i][0] = sz(st) ? st.top() : i;
        st.push(i);
    }
    
    forn(i,N+2) F[i][0] = (H[L[i][0]] < H[R[i][0]] ? R[i][0] : L[i][0]);

    for(int j = 1;j < G;++j) forn(i,N+2) {
        L[i][j] = L[L[i][j-1]][j-1];
        R[i][j] = R[R[i][j-1]][j-1];
        F[i][j] = F[F[i][j-1]][j-1];
    }


}


int minimum_jumps(int A, int B, int C, int D) {
    ++A, ++B, ++C, ++D;
    if(B + 1 == C) {
        return R[B][0] <= D ? 1 : -1;
    }
    int md = B + 1;
    // md is biggest element in (B,C)
    // we have to cross through md eventually
    rforn(i, G) if(R[md][i] < C)md = R[md][i];
    if(R[md][0] > D)return -1;
    if(H[B] > H[md]) {
        return R[B][0] <= D ? 1 : -1;
    }
    int go = B;
    // the optimal is where the H[index] in greatest integer < H[mid] in range [A, B]
    // go is where we start from

    rforn(i, G) if(L[go][i] >= A && H[L[go][i]] < H[md]) go = L[go][i];
    int ret = 0;
    if(L[go][0] >= A) {
        if(R[L[go][0]][0] <= D)return 1;
    }
    else {
        rforn(i, G) if(H[F[go][i]] <= H[md]) {
            go = F[go][i];
            ret += 1 << i;
        }

        if(go == md)return R[go][0] <= D ? ret+1 : -1;
        else if(L[go][0] && R[L[go][0]][0] <= D)return ret+2;
    }

    rforn(i, G) if(R[go][i] < C) {
        ret += 1 << i;
        go = R[go][i];
    }
    go = R[go][0];

    if(C <= go && go <= D)return ret+1;
    else return -1;

}
/*
int main() {

    init(7, {3, 2, 1, 6, 4, 5, 7});
    cout << minimum_jumps(3, 4, 6, 6) << '\n';
}*/

• 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