#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
const int maxn = 1e5 + 100;
int n, k;
int a[maxn];
int t[maxn];
int dp[maxn];
vector<pii> g[maxn];
int l[maxn];
int r[maxn];
int pref[maxn];
void upd(int i, int x){
for(; i <= n; i |= (i + 1)) t[i] = max(t[i], x);
}
int get(int i){
int res = 0;
for(; i > 0; i = (i & (i + 1)) - 1) res = max(res, t[i]);
return res;
}
void init(int N, std::vector<int> H){
n = N;
for(int i = 1; i <= n; i++){
a[i] = H[i-1];
}
k = 1;
while(k < n && a[k+1] > a[k]) k++;
for(int i = k + 1; i <= n; i++){
if(a[i-1] < a[i]) k = 0;
}
for(int i = 1; i <= n; i++){
pref[i] = pref[i-1];
if(a[i] < min(a[i+1], a[i-1])) pref[i]++;
}
}
int max_towers(int L, int R, int D){
if(L == R) return 1;
L++; R++;
if(k){
if(L < k && k < R && max(a[l], a[r]) + D <= a[k]) return 2;
return 1;
}
if(D == 1){
int ans = pref[R-1] - pref[L];
if(a[L] < a[L+1]) ans++;
if(a[R] < a[R-1]) ans++;
return ans;
}
for(int i = 1; i <= n; i++){
t[i] = 0;
g[i].clear();
}
vector<int> v;
for(int i = L; i <= R; i++){
while(v.size() && a[v.back()] <= a[i]) v.pop_back();
v.push_back(i);
l[i] = L - 1;
for(int tl = 0, tr = v.size() - 1; tl <= tr;){
int mid = (tl + tr) >> 1;
if(a[v[mid]] - D < a[i]) tr = mid - 1;
else tl = mid + 1, l[i] = v[mid];
}
}
v.clear();
for(int i = R; i >= L; i--){
while(v.size() && a[v.back()] <= a[i]) v.pop_back();
v.push_back(i);
r[i] = R + 1;
for(int tl = 0, tr = v.size() - 1; tl <= tr;){
int mid = (tl + tr) >> 1;
if(a[v[mid]] - D < a[i]) tr = mid - 1;
else tl = mid + 1, r[i] = v[mid];
}
}
int ans = 0;
for(int i = L; i <= R; i++){
for(auto [p, x]: g[i]){
upd(p, x);
}
dp[i] = get(l[i]) + 1;
g[r[i]].push_back({i, dp[i]});
ans = max(ans, dp[i]);
}
return ans;
}
Compilation message
towers.cpp: In function 'int max_towers(int, int, int)':
towers.cpp:46:29: error: invalid types 'int [100100][int [100100]]' for array subscript
46 | if(L < k && k < R && max(a[l], a[r]) + D <= a[k]) return 2;
| ^
towers.cpp:46:35: error: invalid types 'int [100100][int [100100]]' for array subscript
46 | if(L < k && k < R && max(a[l], a[r]) + D <= a[k]) return 2;
| ^
