| # | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
|---|---|---|---|---|---|---|---|
| 416905 | Theo830 | Rectangles (IOI19_rect) | C++17 | 0 ms | 0 KiB |
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>
using namespace std;
typedef int ll;
const ll INF = 1e9+7;
ll MOD = 998244353;
typedef pair<ll,ll> ii;
#define iii pair<ll,ii>
#define f(i,a,b) for(ll i = a;i < b;i++)
#define pb push_back
#define vll vector<ll>
#define F first
#define S second
#define all(x) (x).begin(), (x).end()
///I hope I will get uprating and don't make mistakes
///I will never stop programming
///sqrt(-1) Love C++
///Please don't hack me
///@TheofanisOrfanou Theo830
///Think different approaches (bs,dp,greedy,graphs,shortest paths,mst)
///Stay Calm
///Look for special cases
///Beware of overflow and array bounds
///Think the problem backwards
///Training
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<ll> distr;
ll rnd(ll a, ll b){return distr(rng)%(b-a+1)+a;}
#include "rect.h"
/*
long long count_rectangles(std::vector<std::vector<int> > a);
int main() {
int n, m;
cin>>n>>m;
vector<vector<int>> a(n, vector<int>(m));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin>>a[i][j];
}
}
long long result = count_rectangles(a);
printf("%lld\n", result);
return 0;
}
*/
bool ok1[700][700][700];
bool ok2[700][700][700];
ll pre1[700][700][701];
ll pre2[700][700][701];
long long count_rectangles(vector<vector<int> > a){
long long ans = 0;
ll n = a.size();
ll m = a[0].size();
f(i,1,n-1){
f(j,1,m-1){
ll Maxi = a[i][j];
ok1[i][j][j] = min(a[i][j-1],a[i][j+1]) > Maxi;
f(k,j+1,m-1){
Maxi = max(Maxi,a[i][k]);
ok1[i][j][k] = min(a[i][j-1],a[i][k+1]) > Maxi;
}
}
}
f(i,1,m-1){
f(j,1,n-1){
ll maxi = a[j][i];
ok2[i][j][j] = min(a[j-1][i],a[j+1][i]) > maxi;
f(k,j+1,n-1){
maxi = max(maxi,a[k][i]);
ok2[i][j][k] = min(a[j-1][i],a[k+1][i]) > maxi;
}
}
}
f(j,1,m-1){
f(jj,j,m-1){
f(k,0,n){
pre1[j][jj][k+1] = pre1[j][jj][k] + ok1[k][j][jj];
}
}
}
f(i,1,n-1){
f(ii,i,n-1){
f(k,0,m){
pre2[i][ii][k+1] = pre2[i][ii][k] + ok2[k][i][ii];
}
}
}
f(i,1,n-1){
f(ii,i,n-1){
f(j,1,m-1){
f(jj,j,m-1){
bool ok = (pre1[j][jj][ii+1] - pre1[j][jj][i]) == (ii - i + 1);
bool ok2 = (pre2[i][ii][jj+1] - pre2[i][ii][j]) == (jj - j + 1);
if(!ok2){
break;
}
ans += ok & ok2;
}
}
}
}
return ans;
}
/*
6 5
4 8 7 5 6
7 4 10 3 5
9 7 20 14 2
9 14 7 3 6
5 7 5 2 7
4 5 13 5 6
*/