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>
#include "rect.h"
using namespace std;
#define fr first
#define sc second
#define pb push_back
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
typedef long long lld;
typedef pair<int, int> pii;
int N, M;
int A[2502][2502];
pii ver[2502][2502], hor[2502][2502];
struct SEGMENT {
int a, b, t, dp;
bool operator < (const SEGMENT &ot)const{
if (t != ot.t) return t < ot.t;
if (a != ot.a) return a < ot.a;
return b < ot.b;
}
bool operator == (const SEGMENT &ot)const{
return a == ot.a && b == ot.b && t == ot.t;
}
bool operator != (const SEGMENT &ot)const{
return a != ot.a || b != ot.b || t != ot.t;
}
};
struct RECT {
int a, b, c, d;
bool operator < (const RECT &ot)const{
if (a != ot.a) return a < ot.a;
if (b != ot.b) return b < ot.b;
if (c != ot.c) return c < ot.c;
return d < ot.d;
}
bool operator != (const RECT &ot)const{
return a != ot.a || b != ot.b || c != ot.c || d != ot.d;
}
};
lld count_rectangles(vector<vector<int> > a)
{
N = sz(a); M = sz(a[0]);
for (int i=1;i<=N;i++) for (int j=1;j<=M;j++) A[i][j] = a[i-1][j-1];
vector <SEGMENT> vertical, horizontal;
for (int i=1;i<=N;i++){
{
stack <int> stk;
for (int j=1;j<=M;j++){
while (!stk.empty() && A[i][stk.top()] <= A[i][j]) stk.pop();
hor[i][j].fr = stk.empty() ? 1 : stk.top()+1;
stk.push(j);
}
}
{
stack <int> stk;
for (int j=M;j;j--){
while (!stk.empty() && A[i][stk.top()] <= A[i][j]) stk.pop();
hor[i][j].sc = stk.empty() ? M : stk.top()-1;
stk.push(j);
}
}
}
for (int j=1;j<=M;j++){
{
stack <int> stk;
for (int i=1;i<=N;i++){
while (!stk.empty() && A[stk.top()][j] <= A[i][j]) stk.pop();
ver[i][j].fr = stk.empty() ? 1 : stk.top()+1;
stk.push(i);
}
}
{
stack <int> stk;
for (int i=N;i;i--){
while (!stk.empty() && A[stk.top()][j] <= A[i][j]) stk.pop();
ver[i][j].sc = stk.empty() ? N : stk.top()-1;
stk.push(i);
}
}
}
for (int i=1;i<=N;i++) for (int j=1;j<=M;j++){
horizontal.pb({hor[i][j].fr, hor[i][j].sc, i});
vertical.pb({ver[i][j].fr, ver[i][j].sc, j});
}
sort(all(horizontal));
sort(all(vertical));
for (auto i=horizontal.begin(),j=i;i!=horizontal.end();i++){
while (*j < SEGMENT{i->a, i->b, i->t-1}) j++;
if (*j == SEGMENT{i->a, i->b, i->t-1}) i->dp = j->dp+1;
else i->dp = 1;
}
for (auto i=vertical.begin(),j=i;i!=vertical.end();i++){
while (*j < SEGMENT{i->a, i->b, i->t-1}) j++;
if (*j == SEGMENT{i->a, i->b, i->t-1}) i->dp = j->dp+1;
else i->dp = 1;
}
vector <RECT> rects;
for (int i=1;i<=N;i++) for (int j=1;j<=M;j++){
int sy = ver[i][j].fr, ey = ver[i][j].sc;
int sx = hor[i][j].fr, ex = hor[i][j].sc;
if (sy == 1 || sx == 1 || ey == N || ex == M) continue;
{
auto it = lower_bound(all(vertical), SEGMENT{sy, ey, ex});
if (it == vertical.end() || *it != SEGMENT{sy, ey, ex} || it->dp <= ex-sx) continue;
}
{
auto it = lower_bound(all(horizontal), SEGMENT{sx, ex, ey});
if (it == horizontal.end() || *it != SEGMENT{sx, ex, ey} || it->dp <= ey-sy) continue;
}
rects.pb({sy, sx, ey, ex});
}
sort(all(rects));
int ans = 0;
for (int i=0;i<sz(rects);i++) if (i == 0 || rects[i-1] != rects[i]) ans++;
return ans;
}
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