이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
#include "rect.h"
#define ll long long
#define ar array
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
using namespace std;
template<typename T> bool ckmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define rint(l, r) uniform_int_distribution<int>(l, r)(rng)
struct fenwick {
int n; vector<int> bit;
fenwick(int a) : n(a), bit(a+1) {}
void update(int pos, int x) {
pos++;
for (int i = pos; i <= n; i += i&-i) bit[i] += x;
}
int sum(int pos) {
int res = 0;
while (pos) res += bit[pos], pos -= pos&-pos;
return res;
}
int query(int l, int r) {
return sum(r+1) - sum(l);
}
};
long long count_rectangles(std::vector<std::vector<int>> a) {
int n = sz(a), m = sz(a[0]);
vector ok(n, vector(n, vector<bool>(m)));
for (int i = 0; i < m; i++) {
for (int j = 1; j < n-1; j++) {
int mx = 0;
for (int k = j; k < n-1; k++) {
ckmax(mx, a[k][i]);
ok[j][k][i] = mx >= min(a[j-1][i], a[k+1][i]);
}
}
}
vector<vector<int>> right(n, vector<int>(m, -1)), left(n, vector<int>(m, m-1));
for (int i = 0; i < n; i++) {
for (int j = 1; j < m-1; j++) {
int mx = 0;
for (int k = j; k < m-1; k++) {
ckmax(mx, a[i][k]);
if (mx >= a[i][j-1]) break;
right[i][j] = k;
}
}
for (int j = m-2; j >= 1; j--) {
int mx = 0;
for (int k = j; k >= 1; k--) {
ckmax(mx, a[i][k]);
if (mx >= a[i][j+1]) break;
left[i][j] = k;
}
}
}
long long ans = 0;
for (int i = 1; i < n-1; i++) {
vector<int> R(m, m-2), L(m, -1);
for (int j = i; j < n-1; j++) {
for (int k = 0; k < m; k++) {
ckmin(R[k], right[j][k]);
ckmax(L[k], left[j][k]);
}
vector<int> to(m, -1);
for (int k = m-1; k >= 0; k--) if (!ok[i][j][k]) {
int l = k;
while (l && !ok[i][j][l-1]) l--;
for (int z = l; z <= k; z++) to[z] = k;
k = l;
}
fenwick fw(m);
for (int k = 1; k < m-1; k++) {
int r = min(to[k], R[k]);
if (r >= k) fw.update(r, 1), fw.update(k-1, -1);
if (L[k] <= k) ans += fw.query(L[k], k);
}
}
}
return ans;
}
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