Submission #931614

#	Time	Username	Problem	Language	Result	Execution time	Memory
931614		Arp	밀림 점프 (APIO21_jumps)	C++17	100 / 100	820 ms	63096 KiB

jumps

#include <bits/stdc++.h>
 
using namespace std;
 
const int N = 2e5+10;
const int K = 18;
const int INF = 1e9;
 
int up[K][N], mvr[K][N], mvl[K][N], h[N], idx[N], lh[N], rh[N], n;
int sp[K][N], lg[N];
 
int que(int l, int r) {
    int i=lg[r-l+1];
    return max(sp[i][l], sp[i][r-(1<<i)+1]);
}
 
void init(int N, std::vector<int> H) {
    n=N;
    for (int i=0; i<n; ++i) h[i+1]=H[i], idx[h[i+1]]=i+1;
 
    h[0]=INF-1, h[n+1]=INF;
    stack<int> st; st.push(0);
 
    for (int i=1; i<=n; ++i) {
        while (h[st.top()] <= h[i]) st.pop();
        assert(!st.empty());
        lh[i]=st.top(), st.push(i);
    } st=stack<int>();
 
    for (int i=1; i<=n; ++i) sp[0][i]=h[i];
    for (int j=1; j<K; ++j) {
        for (int i=1; i+(1<<(j-1))<=n; ++i) sp[j][i]=max(sp[j-1][i], sp[j-1][i+(1<<(j-1))]);
    }
 
    lg[1]=0;
    for (int i=2; i<N; ++i) lg[i]=lg[i/2]+1;
 
    st.push(n+1);
    for (int i=n; i>=1; --i) {
        while (h[st.top()] <= h[i]) st.pop();
        assert(!st.empty());
        rh[i]=st.top(), st.push(i);
    }
 
    // assume A=B, C=D for testing reasons
    for (int i=n; i>=1; --i) {
        int j=idx[i];
 
        int r=rh[j], l=lh[j];
        assert(h[l] > h[j] && h[j] < h[r]);
 
        if (h[l] > h[r] && l > 0) up[0][j]=l;
        else up[0][j]=r;
 
        for (int k=1; k<K; ++k) up[k][j]=up[k-1][up[k-1][j]];
    }
 
    for (int k=0; k<K; ++k) mvr[k][n+1]=n+1;
 
    for (int i=n; i>=1; --i) {
        int j=idx[i];
        mvr[0][j]=rh[j];
        for (int k=1; k<K; ++k) mvr[k][j]=mvr[k-1][mvr[k-1][j]];
    }
 
    for (int i=n; i>=1; --i) {
        int j=idx[i];
        mvl[0][j]=lh[j];
        for (int k=1; k<K; ++k) mvl[k][j]=mvl[k-1][mvl[k-1][j]];
    }
}
 
int minimum_jumps(int a, int b, int c, int d) {
    ++a, ++b, ++c, ++d;
 
    int x=b, mxv=que(c, d), ans=0;
    for (int j=K-1; j>=0; --j) {
        int u=mvl[j][x];
        if (u >= a && h[u] < mxv) x=u;
    }
 
    // find greatest A such that A < c and rh[A] >= c and h[A] < h[C]
    for (int j=K-1; j>=0; --j) {
        int u=up[j][x];
        if (u < c && rh[u] < c && h[u] < mxv) x=u, ans+=(1<<j);
    }
 
    if (up[0][x] < c && h[up[0][x]] < mxv && rh[x] < c) x=up[0][x], ++ans;
 
    for (int j=K-1; j>=0; --j) {
        int u=mvr[j][x];
        if (u < c) x=u, ans+=(1<<j);
    }
    if (rh[x] >= c && rh[x] <= d) return ans+1;
    return -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