제출 #344383

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
344383		nicolaalexandra	최솟값 배열 (IZhO11_hyper)	C++14	100 / 100	1263 ms	62404 KiB

hyper

#include <bits/stdc++.h>
#define DIM 40
#define INF 2000000000
using namespace std;

int rmq[7][DIM][DIM][DIM][DIM];
int n,m,i,j,k,t,p,mask;

int main (){

    //ifstream cin ("date.in");
    //ofstream cout ("date.out");

    cin>>n>>m;
    for (i=1;i<=n;i++)
        for (j=1;j<=n;j++)
            for (k=1;k<=n;k++)
                for (t=1;t<=n;t++)
                    cin>>rmq[0][i][j][k][t];

    for (p=1;(1<<p)<=m;p++){

        for (i=1;i<=n;i++)
            for (j=1;j<=n;j++)
                for (k=1;k<=n;k++)
                    for (t=1;t<=n;t++){
                        rmq[p][i][j][k][t] = INF;
                        int val = (1<<(p-1));
                        for (mask=0;mask<16;mask++){
                            int ii = i, jj = j, kk = k, tt = t;
                            if (mask&1)
                                ii += val;
                            if (mask&2)
                                jj += val;
                            if (mask&4)
                                kk += val;
                            if (mask&8)
                                tt += val;

                            if (ii <= n && jj <= n && kk <= n && tt <= n)
                                rmq[p][i][j][k][t] = min (rmq[p][i][j][k][t],rmq[p-1][ii][jj][kk][tt]);
                        }}}
    p--;

    for (i=1;i<=n-m+1;i++)
        for (j=1;j<=n-m+1;j++)
            for (k=1;k<=n-m+1;k++)
                for (t=1;t<=n-m+1;t++){
                    int mini = INF, val = (1<<p);
                    for (mask=0;mask<16;mask++){
                        int ii = i, jj = j, kk = k, tt = t;
                        if (mask&1)
                            ii += m - val;
                        if (mask&2)
                            jj += m - val;
                        if (mask&4)
                            kk += m - val;
                        if (mask&8)
                            tt += m - val;

                        if (ii <= n && jj <= n && kk <= n && tt <= n)
                            mini = min (mini,rmq[p][ii][jj][kk][tt]);
                    }
                    cout<<mini<<" ";
                }


    return 0;
}

• Subtask 1: 100 / 100