#include "rect.h"
#include <bits/stdc++.h>
using namespace std;
const int SMALL = -2;
const int BIG = 1234567890;
struct ordered_set { // This is actually just a fenwick tree
vector<int> nums;
int size = 4096;
int ct = 0;
ordered_set() {
nums.resize(size+1);
}
void insert(int n) {
ct++;
n++;
while(n <= size) {
nums[n]++;
n += (n & (-n));
}
}
void erase(int n) {
ct--;
n++;
while(n <= size) {
nums[n]--;
n += (n & (-n));
}
}
int count(int n) {
int ans = 0;
while(n) {
ans += nums[n];
n -= (n & (-n));
}
return ct - ans;
}
};
long long count_rectangles(vector<vector<int>> a) {
// Idea: consider "good" segments, where the ones to the left and to the right are bigger
// It turns out there are O(n) such subsegments in a segment with n elements
// Thus, in a "good" rectangle, each row and column must be a good segment
// Fix the upper left corner of a rectangle, consider some particular segment
// which starts at this square, WLOG it is horizontal
// The number of rectangles that "use" this segment is equal to the number of
// vertical segments that all have the same length, which cover the same width
// You can do this with like, sweep line
int n = a.size();
int m = a[0].size();
if(min(n, m) <= 2) return 0;
vector<vector<vector<pair<int, int>>>> ps(n);
vector<pair<int, int>> v;
for(int i = 0; i < n; i++) {
ps[i].resize(m);
v.push_back({BIG, -1});
for(int j = 0; j < m; j++) {
while(v.back().first < a[i][j]) {
if(j - v.back().second > 1) ps[i][v.back().second+1].push_back({1, j-1});
while(v.size() >= 2) {
if(v.back().first == v[v.size()-2].first) v.pop_back();
else break;
}
v.pop_back();
}
if(j - v.back().second > 1 && v.size() > 1) ps[i][v.back().second+1].push_back({1, j-1});
v.push_back({a[i][j], j});
}
vector<pair<int, int>>().swap(v);
}
for(int i = 0; i < m; i++) {
v.push_back({BIG, -1});
for(int j = 0; j < n; j++) {
while(v.back().first < a[j][i]) {
if(j - v.back().second > 1) ps[v.back().second+1][i].push_back({-1, j-1});
while(v.size() >= 2) {
if(v.back().first == v[v.size()-2].first) v.pop_back();
else break;
}
v.pop_back();
}
if(j - v.back().second > 1 && v.size() > 1) ps[v.back().second+1][i].push_back({-1, j-1});
v.push_back({a[j][i], j});
}
vector<pair<int, int>>().swap(v);
}
for(int i = 0; i < n; i++) {
for(int j = 0; j < m; j++) {
for(auto &p : ps[i][j]) {
swap(p.first, p.second);
// cout << i << ' ' << j << ' ' << p.first << ' ' << p.second << '\n';
}
sort(ps[i][j].begin(), ps[i][j].end());
}
}
for(int i = n-2; i > 0; i--) {
for(int j = m-2; j > 0; j--) {
for(auto &p : ps[i][j]) {
if(p.second > 0) { // strokes horizontal
pair<int, int> pe = {p.first, 0};
auto it = lower_bound(ps[i+1][j].begin(), ps[i+1][j].end(), pe);
if(it != ps[i+1][j].end()) {
auto pa = (*it);
// cout << pa.first << ' ' << pa.second << '\n';
if(pa.first == p.first && pa.second > 0) {
p.second += pa.second;
}
}
}
else { // strokes vertical
pair<int, int> pe = {p.first, -BIG};
auto it = lower_bound(ps[i][j+1].begin(), ps[i][j+1].end(), pe);
if(it != ps[i][j+1].end()) {
auto pa = (*it);
if(pa.first == p.first && pa.second < 0) {
p.second += pa.second;
}
}
}
}
}
}
long long ans = 0;
ordered_set s;
for(int i = 1; i < n-1; i++) {
for(int j = 1; j < m-1; j++) {
vector<pair<pair<int, int>, int>> pts;
for(auto p : ps[i][j]) {
// cout << i << ' ' << j << ' ' << p.first << ' ' << p.second << '\n';
if(p.second > 0) pts.push_back({{i-1+p.second, -p.first}, 1});
else pts.push_back({{p.first, -j+1+p.second}, -1});
}
sort(pts.begin(), pts.end());
for(auto p : pts) {
// cout << p.first.first << ' ' << p.first.second << ' ' << p.second << '\n';
if(p.second == 1) {
ans += s.count(abs(p.first.second));
}
else {
s.insert(abs(p.first.second));
}
}
for(auto p : pts) {
if(p.second == -1) {
s.erase(abs(p.first.second));
}
}
// cout << ans << '\n';
}
}
return ans;
}
