#include "jumps.h"
#include <bits/stdc++.h>
using namespace std;
// st0[u][z] = st0[st0[u][z-1]][z-1];
#include <vector>
vector<vector<int>> adj;
vector<vector<int>> radj;
int n = 0;
vector<vector<int>> st0;
vector<vector<int>> st1;
vector<int> visited;
vector<int> h;
// SEGMENT TREE
const int S_ = 262144 * 4 + 5;
vector<vector<int>> tree(S_, vector<int>());
// fast[l/r][node][number] = biggest element <= number in l/r subtree of node
// wait NVM
// For each element inside a segment tree node, we need the O(1) function for that node
//vector<vector<map<int, int>>> fast(2, vector<map<int, int>>(262144 * 2 + 5, map<int, int>()));
// fast[l/r][node][index within that node] = biggest element <= node[index] in l/r subtree of node
vector<vector<vector<int>>> fast(2, vector<vector<int>>(262144 * 4 + 5, vector<int>()));
vector<vector<int>> RMQ;
/*
THE PLAN
To find the best tree in the range [A, B]
- Find the rightmost tree in the range that is taller than C, call it X using a
Segtree that supports range rightmost threshold queries!
Then a segtree that supports range upper bound queries!
A != B but C = D
A ... B
Let X = rightmost tree between A and B that is taller than C
Now use range upper bound to find tallest tree between X and B that is shorter than C
And then do RMQ(C) for the tree we want to start at
Then we need to find the fastest way to get to (C -- D)
Solve the problem in reverse
- Then do the range upper bound query within the range [X, B]
*/
//vector<vector<vector<int>>> fast(2, vector<vector<int>>(262144 * 2 + 5, vector<int>()));
//int[2][262144 * 2 +5][] fast;
// For each node in the segment tree, we have O(log(n)) numbers
// n logn + n logn + n logn ... logn times
int fast_log(int x){
return x ? __builtin_clzll(1) - __builtin_clzll(x) : -1;
}
void build(int j, int tl, int tr) {
if (tl == tr) {
tree[j] = {h[tl]};
return;
} else {
int tm = (tl + tr) / 2;
build(j*2, tl, tm);
build(j*2+1, tm+1, tr);
merge(tree[j * 2].begin(), tree[j * 2].end(), tree[j * 2 + 1].begin(), tree[j * 2 + 1].end(), back_inserter(tree[j]));
int pl = 0;
int pr = 0;
for(int x = 0; x < tree[j].size(); x ++){
if(tree[j * 2][pl] == tree[j][x]){
fast[0][j].push_back(pl);
if(x == 0){ fast[1][j].push_back(-1);
} else { fast[1][j].push_back(fast[1][j][fast[1][j].size() - 1]); }
pl += 1;
} else {
fast[1][j].push_back(pr);
if(x == 0){ fast[0][j].push_back(-1);
} else { fast[0][j].push_back(fast[0][j][fast[0][j].size() - 1]); }
pr += 1;
}
}
if(fast[1][j].size() == 0) fast[1][j].push_back(-1);
if(fast[0][j].size() == 0) fast[0][j].push_back(-1);
}
}
vector<int> query;
int rmq0(int l, int r){
int i = fast_log(r - l + 1);
return max(RMQ[i][l], RMQ[i][r - (1 << i) + 1]);
}
int qu = 0;
void rightmost(int threshold, int ql, int qr, int lbound, int rbound, int j = 1){
int mid = (lbound + rbound) / 2;
if(lbound == rbound){
qu = max(qu, lbound);
return;
}
int call = 0;
if(rmq0(mid+1, rbound) >= threshold){
if(ql <= rbound and qr >= mid + 1){
call = 1;
rightmost(threshold, ql, qr, mid+1, rbound, j * 2 + 1);
}
}
if(rmq0(lbound, mid) >= threshold){
if(!call){
if(ql <= mid and qr >= lbound){
rightmost(threshold, ql, qr, lbound, mid, j * 2);
}
}
}
}
void rangeq(int bound, int ql, int qr, int lbound, int rbound, int j = 1){
if(qr < lbound || rbound < ql){
//return INT_MAX;
// DONT DO ANYTHING
} else if(ql <= lbound and rbound <= qr){
if(bound >= 0){
query.push_back(tree[j][bound]);
}
} else {
int mid = (lbound + rbound) / 2;
if(bound >= 0){
rangeq(fast[0][j][bound], ql, qr, lbound, mid, j * 2);
rangeq(fast[1][j][bound], ql, qr, mid+1, rbound, j * 2 + 1);
}
}
}
// Construct sparse tables
void dfs0(int u, int v){
if(!visited[u]){
visited[u] = 1;
st0[u][0] = v;
for(int z = 1; z <= ceil(log2(n)); z ++){
st0[u][z] = st0[st0[u][z-1]][z-1];
}
for(int i = 0; i < radj[u].size(); i ++){
int highedge = 1;
for(int j = 0; j < adj[radj[u][i]].size(); j ++){
if(h[adj[radj[u][i]][j]] > h[u]){
highedge = 0;
}
}
if(highedge){
dfs0(radj[u][i], u);
}
}
}
}
void dfs1(int u, int v){
if(!visited[u]){
visited[u] = 1;
st1[u][0] = v;
for(int z = 1; z <= ceil(log2(n)); z ++){
st1[u][z] = st1[st1[u][z-1]][z-1];
}
for(int i = 0; i < radj[u].size(); i ++){
int lowedge = 1;
for(int j = 0; j < adj[radj[u][i]].size(); j ++){
if(h[adj[radj[u][i]][j]] < h[u]){
lowedge = 0;
}
}
if(lowedge){
dfs1(radj[u][i], u);
}
}
}
}
int subtask1 = 1;
vector<int> htt;
void init(int N, vector<int> H) {
n = N;
h = H;
stack<pair<int, int>> s;
vector<vector<int>> adj0(N, vector<int>());
int highest = 0;
int idx = 0;
vector<vector<int>> radj0(N, vector<int>());
// Monotonic stack to construct adjacency list in O(n)
for (int i = 0; i < N; i ++) {
if(i > 0){
if(H[i] != i + 1){
subtask1 = 0;
}
}
while (!s.empty() && s.top().first < H[i]){ s.pop(); }
if (!s.empty()){
adj0[i].push_back(s.top().second);
radj0[s.top().second].push_back(i);
}
s.push({H[i], i});
if(H[i] > highest){
highest = H[i];
idx = i;
}
}
while(!s.empty()){ s.pop(); }
for (int i = N - 1; i >= 0; i --) {
while (!s.empty() && s.top().first < H[i]){ s.pop(); }
if (!s.empty()){
adj0[i].push_back(s.top().second);
radj0[s.top().second].push_back(i);
}
s.push({H[i], i});
}
adj = adj0;
radj = radj0;
vector<vector<int>> st(N, vector<int>(ceil(log2(n)) + 1, 0));
vector<vector<int>> rmqt(ceil(log2(n))+1, vector<int>(N, 0));
st0 = st;
st1 = st;
RMQ = rmqt;
for(int j = 0; j < n; j ++){
RMQ[0][j] = H[j];
}
for(int i = 1; i <= ceil(log2(n)); i ++){
for(int j = 0; j < n; j ++){
RMQ[i][j] = max(RMQ[i-1][j], RMQ[i-1][j + (1 << (i - 1))]);
}
}
vector<int> v(N, 0);
visited = v;
htt = v;
for(int i = 0; i < H.size(); i ++){
htt[H[i] - 1] = i;
}
dfs0(idx, idx);
visited = v;
dfs1(idx, idx);
build(1, 0, N + 1);
}
int minimum_jumps(int A, int B, int C, int D){
qu = -1;
rightmost(h[C], A, B, 0, n);
//return rmq0(A, B);
int start = 0;
if(qu != -1){
start = rmq0(qu + 1, B);
} else {
start = rmq0(A, B);
}
//int start = rmq0(max(qu+1, A), B);
//return qu;
if(C == D){
int jumps = 0;
int u = htt[start - 1];
for (int i = ceil(log2(n)); i >= 0; --i) {
if(h[st0[u][i]] < h[C]){
u = st0[u][i];
jumps += (1 << i);
}
}
// If the first sparse table is all we need, we're done
if(st0[u][0] == C){
return jumps + 1;
} else {
// Otherwise start at the next sparse table
for (int i = ceil(log2(n)); i >= 0; --i) {
if(h[st1[u][i]] < h[C]){
u = st1[u][i];
jumps += (1 << i);
}
}
if(st1[u][0] == C){
return jumps + 1;
}
}
return -1;
} /*{
if(subtask1){
return C - B;
} else {
queue<pair<int, int>> q;
for(int i = A; i <= B; i ++){
q.push({0, i});
}
int node = 0; int dist = 0;
vector<int> vi(n, 0);
while(!q.empty()){
node = q.front().second;
dist = q.front().first;
//cout << node << "\n";
q.pop();
if(node >= C && node <= D){
return dist;
//break;
}
if(!vi[node]){
vi[node] = 1;
for(int i = 0; i < adj[node].size(); i ++){
q.push({dist + 1, adj[node][i]});
}
}
}
return -1;
}
}*/
return -1;
}
/*
The rightmost tree within range that is taller than
g++ -std=gnu++17 -O2 -pipe -o jumps jumps.cpp stub.cpp
7 3
3 7 5 6 4 2 1
0 1 6 6
0 2 6 6
0 3 6 6
0 4 4 4
0 5 4 4
0 6 4 4
3 6 5 5
4 6 5 5
5 6 5 5
6 6 6 6
if(A == B and C == D){
int jumps = 0;
int u = A;
for (int i = ceil(log2(n)); i >= 0; --i) {
if(h[st0[u][i]] < h[C]){
u = st0[u][i];
jumps += (1 << i);
}
}
// If the first sparse table is all we need, we're done
if(st0[u][0] == C){
return jumps + 1;
} else {
// Otherwise start at the next sparse table
for (int i = ceil(log2(n)); i >= 0; --i) {
if(h[st1[u][i]] < h[C]){
u = st1[u][i];
jumps += (1 << i);
}
}
if(st1[u][0] == C){
return jumps + 1;
}
}
return -1;
// Otherwise, if it's subtask 1, we do our dumb solution
} else if(subtask1){
return C - B;
// Subtask 6 - wavelet tree
} else if(C == D){
cout << 100;
query.clear();
//rangeq(C, A, B, 0, n);
for(int i = 0; i < query.size(); i ++){
cout << query[i] << " ";
} cout << "\n";
// Now it's only subtask 2, 3 or 4 so we do BFS
} else {
//return u;
if(jumps){
return jumps + 1;
} else {
return 0;
}
queue<pair<int, int>> q;
for(int i = A; i <= B; i ++){
q.push({0, i});
}
int node = 0; int dist = 0;
vector<int> vi(n, 0);
while(!q.empty()){
node = q.front().second;
dist = q.front().first;
//cout << node << "\n";
q.pop(); x
if(node >= C && node <= D){
return dist;
//break;
}
if(!vi[node]){
vi[node] = 1;
for(int i = 0; i < adj[node].size(); i ++){
q.push({dist + 1, adj[node][i]});
}
}
}
return -1;
}
*/
