Submission #58118

#	Time	Username	Problem	Language	Result	Execution time	Memory
58118		Benq	Hyper-minimum (IZhO11_hyper)	C++14	100 / 100	484 ms	43824 KiB

hyper

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N,M, X[40][40][40][40];

void solve(int d) {
    if (d == 3) {
        F0R(i,N) F0R(j,N) F0R(k,N) {
            deque<pi> d;
            F0Rd(l,N) {
                while (sz(d) && d.back().f >= X[i][j][k][l]) d.pop_back();
                d.pb({X[i][j][k][l],l});
                if (l <= N-M) {
                    while (d.front().s >= l+M) d.pop_front();
                    X[i][j][k][l] = d.front().f;
                }
            }
        }
    } else if (d == 2) {
        F0R(i,N) F0R(j,N) F0R(l,N-M+1) {
            deque<pi> d;
            F0Rd(k,N) {
                while (sz(d) && d.back().f >= X[i][j][k][l]) d.pop_back();
                d.pb({X[i][j][k][l],k});
                if (k <= N-M) {
                    while (d.front().s >= k+M) d.pop_front();
                    X[i][j][k][l] = d.front().f;
                }
            }
        }
    } else if (d == 1) {
    	F0R(i,N) F0R(k,N-M+1) F0R(l,N-M+1) {
            deque<pi> d;
            F0Rd(j,N) {
                while (sz(d) && d.back().f >= X[i][j][k][l]) d.pop_back();
                d.pb({X[i][j][k][l],j});
                if (j <= N-M) {
                    while (d.front().s >= j+M) d.pop_front();
                    X[i][j][k][l] = d.front().f;
                }
            }
        }
    } else {
    	F0R(j,N-M+1) F0R(k,N-M+1) F0R(l,N-M+1) {
            deque<pi> d;
            F0Rd(i,N) {
                while (sz(d) && d.back().f >= X[i][j][k][l]) d.pop_back();
                d.pb({X[i][j][k][l],i});
                if (i <= N-M) {
                    while (d.front().s >= i+M) d.pop_front();
                    X[i][j][k][l] = d.front().f;
                }
            }
        }
    }
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> M;
    F0R(i,N) F0R(j,N) F0R(k,N) F0R(l,N) cin >> X[i][j][k][l];
    F0Rd(i,4) solve(i);
    F0R(i,N-M+1) F0R(j,N-M+1) F0R(k,N-M+1) F0R(l,N-M+1) cout << X[i][j][k][l] << " ";
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 100 / 100