Submission #416911

#	Time	Username	Problem	Language	Result	Execution time	Memory
416911		Theo830	Rectangles (IOI19_rect)	C++17	8 / 100	2867 ms	61820 KiB

rect

#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][70][700];
bool ok2[700][70][700];
ll pre1[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(j,1,m-1){
                ll posa = 0;
                f(jj,j,m-1){
                    posa += ok2[jj][i][ii];
                    bool ok = (pre1[j][jj][ii+1] - pre1[j][jj][i]) == (ii - i  + 1);
                    bool ok2 = (posa == (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
*/

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