이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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 (x >= c && x <= d) return ans;
if (rh[x] >= c && rh[x] <= d) return ans+1;
return -1;
}
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