This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
#include "jumps.h"
using namespace std;
template<typename T> void ono_min(T &MIN, T CMP) { if(MIN > CMP) MIN = CMP; }
template<typename T> void ono_max(T &MAX, T CMP) { if(MAX < CMP) MAX = CMP; }
typedef pair<int, int> pii;
#define ff first
#define ss second
#define MP make_pair
const int mxk = 32;
const int mxn = 2e5 + 10;
int L[mxn], R[mxn], n;
vector<int> adj[mxn];
int dis[mxn], h[mxn];
int mx[mxk][mxn];
int mn[mxk][mxn];
bool mx_valid(int k, int i) { return 1 <= mx[k][i] && mx[k][i] <= n; }
bool mn_valid(int k, int i) { return 1 <= mn[k][i] && mn[k][i] <= n; }
void init(int N, vector<int> H) {
memset(L, -1, sizeof L);
memset(R, -1, sizeof R);
n = N;
stack<int> st; st.push(0);
for(int i = 1; i < N; i++) {
while(st.size() && H[st.top()] <= H[i])
st.pop();
if(st.size())
L[i] = st.top();
st.push(i);
}
stack<int> en; en.push(0);
for(int i = N - 1; i >= 0; i--) {
while(en.size() && H[en.top()] <= H[i])
en.pop();
if(en.size())
R[i] = en.top();
en.push(i);
}
for(int i = 0; i < N; i++) {
h[i] = H[i];
if(L[i] > -1) {
adj[i].push_back(L[i]);
ono_max(mx[0][H[i]], H[L[i]]);
if(mn[0][H[i]] == 0)
mn[0][H[i]] = H[L[i]];
else
ono_min(mn[0][H[i]], H[L[i]]);
}
if(R[i] > -1) {
adj[i].push_back(R[i]);
ono_max(mx[0][H[i]], H[R[i]]);
if(mn[0][H[i]] == 0)
mn[0][H[i]] = H[R[i]];
else
ono_min(mn[0][H[i]], H[R[i]]);
}
}
for(int k = 1; k < mxk; k++)
for(int i = 1; i <= n; i++) {
int mnj = mn[k - 1][i];
int mxj = mx[k - 1][i];
if(1 <= mnj && mnj <= n) mn[k][i] = mn[k - 1][mnj];
if(1 <= mxj && mxj <= n) mx[k][i] = mx[k - 1][mxj];
}
}
int minimum_jumps(int A, int B, int C, int D) {
if(A != B) exit(1);
if(C != D) exit(1);
if(h[A] > h[C]) return -1;
int s = h[A], f = h[C], ans = 0;
for(int k = mxk - 1; k >= 0; k--)
if(1 <= mx[k][s] && mx[k][s] <= n && mx[k][s] <= f) {
s = mx[k][s];
ans ^= (1 << k);
}
for(int k = mxk - 1; k >= 0; k--)
if(1 <= mn[k][s] && mn[k][s] <= n && mn[k][s] <= f) {
s = mn[k][s];
ans ^= (1 << k);
}
if(f == s) return ans;
return -1;
}
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