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>
using namespace std;
#define ll long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define ff first
#define ss second
#define pb push_back
#define SZ(x) ((int)(x).size())
#define all(x) x.begin(), x.end()
#define debug(x) cout << #x << ": " << x << " "
#define nl cout << "\n"
#define rep(i, a, b) for(int i = (a); i <= (b); i++)
#define per(i, a, b) for(int i = (a); i >= (b); i--)
//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long count_rectangles(vector<vector<int>> a){
int N = SZ(a), M = SZ(a[0]);
vector<int> rt[N][M], dw[N][M];
rep(i, 0, N - 1){
vector<int> st;
per(j, M - 1, 0){
while(!st.empty() && a[i][j] > a[i][st.back()]){
if(j + 1 < st.back()) rt[i][j].pb(st.back());
st.pop_back();
}
if(!st.empty() && j + 1 < st.back()) rt[i][j].pb(st.back());
if(!st.empty() && a[i][j] == a[i][st.back()]) st.pop_back();
st.pb(j);
}
}
rep(j, 0, M - 1){
vector<int> st;
per(i, N - 1, 0){
while(!st.empty() && a[i][j] > a[st.back()][j]){
if(i + 1 < st.back()) dw[i][j].pb(st.back());
st.pop_back();
}
if(!st.empty() && i + 1 < st.back()) dw[i][j].pb(st.back());
if(!st.empty() && a[i][j] == a[st.back()][j]) st.pop_back();
st.pb(i);
}
}
ll ans = 0;
vector<pii> vt[M][M];
rep(i, 0, N - 1){
rep(j, 0, M - 1){
for(int j2 : rt[i][j]){
if(SZ(vt[j][j2]) && vt[j][j2].back().ss == i - 1)
vt[j][j2].back().ss = i;
else vt[j][j2].pb({i, i});
}
}
}
rep(i, 1, N - 2){
rep(j, 0, M - 1){
int curj = j + 1;
int st[2500], barr[2500]; int stsz = 0, barrsz;
for(int i2 : dw[i - 1][j + 1]) st[stsz++] = i2;
for(int j2 : rt[i][j]){
if(!stsz) break;
while(curj < j2){
barrsz = 0;
int a = 0, b = 0;
while(a < SZ(dw[i - 1][curj]) && b < stsz){
if(dw[i - 1][curj][a] == st[b]){
barr[barrsz++] = st[b]; a++, b++;
} else if(dw[i - 1][curj][a] < st[b]) a++;
else b++;
}
swap(st, barr);
stsz = barrsz;
curj++;
}
int a = 0, b = 0;
while(a < SZ(vt[j][j2]) && b < stsz){
int i2 = st[b];
if(vt[j][j2][a].ff <= i && i2 - 1 <= vt[j][j2][a].ss) ans++, b++;
else if(vt[j][j2][a].ss < i2 - 1) a++;
else b++;
}
}
}
}
return ans;
}
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