제출 #337757

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
337757		tengiz05	최솟값 배열 (IZhO11_hyper)	C++17	100 / 100	347 ms	52460 KiB

hyper

#include <bits/stdc++.h>
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define pii pair<int, int>
#define ff first
#define ss second
#define PI acos(-1)
#define ld long double
const int mod = 1e9+7, N = 36;
int msb(int val){return sizeof(int)*8-__builtin_clzll(val);}
int n, m;
int ans[5][N][N][N][N];

void solve(int test_case){
	int i, j, k, l;
	cin >> n >> m;
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
			for(k=1;k<=n;k++)
				for(l=1;l<=n;l++)
					cin >> ans[4][i][j][k][l];
	
	
	int li = n-m+1;
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
			for(k=1;k<=n;k++)
				for(l=1;l<=n;l++){
					ans[3][i][j][k][l] = mod;
					for(int t=l;t<=min(n,l+m-1);t++){
						ans[3][i][j][k][l] = min(ans[3][i][j][k][l], ans[4][i][j][k][t]);
					}
				}
				
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
			for(k=1;k<=n;k++)
				for(l=1;l<=n;l++){
					ans[2][i][j][k][l] = mod;
					for(int t=k;t<=min(n,k+m-1);t++){
						ans[2][i][j][k][l] = min(ans[2][i][j][k][l], ans[3][i][j][t][l]);
					}
				}
	
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
			for(k=1;k<=n;k++)
				for(l=1;l<=n;l++){
					ans[1][i][j][k][l] = mod;
					for(int t=j;t<=min(n,j+m-1);t++){
						ans[1][i][j][k][l] = min(ans[1][i][j][k][l], ans[2][i][t][k][l]);
					}
				}
	
	for(i=1;i<=n;i++)
		for(j=1;j<=n;j++)
			for(k=1;k<=n;k++)
				for(l=1;l<=n;l++){
					ans[0][i][j][k][l] = mod;
					for(int t=i;t<=min(n,i+m-1);t++){
						ans[0][i][j][k][l] = min(ans[0][i][j][k][l], ans[1][t][j][k][l]);
					}
				}
	
	
	for(i=1;i<=li;i++)
		for(j=1;j<=li;j++)
			for(k=1;k<=li;k++)
				for(l=1;l<=li;l++)
					cout << ans[0][i][j][k][l] << ' ';
	
	cout << '\n';
	return;
}

signed main(){
	FASTIO;
#define MULTITEST 0
#if MULTITEST
	int ___T;
	cin >> ___T;
	for(int T_CASE = 1; T_CASE <= ___T; T_CASE++)
		solve(T_CASE);
#else
	solve(1);
#endif
	return 0;
}

• Subtask 1: 100 / 100