Submission #416867

#	Time	Username	Problem	Language	Result	Execution time	Memory
416867		Theo830	Rectangles (IOI19_rect)	C++17	25 / 100	5046 ms	1048580 KiB

rect

#include <bits/stdc++.h>
using namespace std;
typedef long long 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;
}
*/
long long count_rectangles(vector<vector<int> > a){
    ll ans = 0;
    ll n = a.size();
    ll m = a[0].size();
    ll Maxi[n][m][m];
    ll maxi[m][n][n];
    f(i,1,n-1){
        f(j,1,m-1){
            Maxi[i][j][j] = a[i][j];
            f(k,j+1,m-1){
                Maxi[i][j][k] = max(Maxi[i][j][k-1],1LL * a[i][k]);
            }
        }
    }
    f(i,1,m-1){
        f(j,1,n-1){
            maxi[i][j][j] = a[j][i];
            f(k,j+1,n-1){
                maxi[i][j][k] = max(maxi[i][j][k-1],1LL * a[k][i]);
            }
        }
    }
    f(i,1,n-1){
        f(j,1,m-1){
            f(ii,i,n-1){
                f(jj,j,m-1){
                    bool ok = 1;
                    f(k,i,ii+1){
                        ok &= (Maxi[k][j][jj] < min(a[k][j-1],a[k][jj+1]));
                    }
                    f(k,j,jj+1){
                        ok &= (maxi[k][i][ii] < min(a[i-1][k],a[ii+1][k]));
                    }
                    ans += ok;
                }
            }
        }
    }
	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
*/

• Subtask 1: 8 / 8
• Subtask 2: 7 / 7
• Subtask 3: 0 / 12
• Subtask 4: 0 / 22
• Subtask 5: 10 / 10
• Subtask 6: 0 / 13
• Subtask 7: 0 / 28