Submission #578848

#	Time	Username	Problem	Language	Result	Execution time	Memory
578848		Theo830	Rectangles (IOI19_rect)	C++17	37 / 100	1167 ms	1048576 KiB

rect

    #pragma GCC optimize("Ofast")
    #pragma GCC optimize ("O3")
    #pragma GCC optimization ("unroll-loops")
    #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
    #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;
    }
    */
    long long count_rectangles(vector<vector<int> > a){
        ll ans = 0;
        ll n = a.size();
        ll m = a[0].size();
        bool ok1[n][m][m];
        bool ok2[m][n][n];
        memset(ok1,0,sizeof ok1);
        memset(ok2,0,sizeof ok2);
        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;
                }
            }
        }
        ll pre1[m][m][n+1];
        memset(pre1,0,sizeof pre1);
        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];
                }
            }
        }
        ll pre2[n][n][m+1];
        memset(pre2,0,sizeof pre2);
        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
    */

Compilation message (stderr)

rect.cpp:5: warning: ignoring '#pragma GCC optimization' [-Wunknown-pragmas]
    5 |     #pragma GCC optimization ("unroll-loops")
      |

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