제출 #13260

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
13260		baneling100	최솟값 배열 (IZhO11_hyper)	C++98	100 / 100	499 ms	34392 KiB

hyper

#include <stdio.h>
#include <algorithm>
#include <deque>
#define N_MAX 35
#define INF 1000000001

using namespace std;

deque <int> Deq;
int N, M, X[4][N_MAX+1][N_MAX+1][N_MAX+1][N_MAX+1], Y[N_MAX+1][N_MAX+1][N_MAX+1][N_MAX+1];

void input(void) {

    int i, j, k, l;

    scanf("%d %d",&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++) {
                    scanf("%d",&X[0][i][j][k][l]);
                    X[1][i][j][k][l]=X[2][i][j][k][l]=X[3][i][j][k][l]=Y[i][j][k][l]=INF;
                }
}

void process(void) {

    int i, j, k, l;

    for(j=1 ; j<=N ; j++)
        for(k=1 ; k<=N ; k++)
            for(l=1 ; l<=N ; l++) {
                X[0][0][j][k][l]=INF;
                Deq.clear();
                for(i=1 ; i<=M-1 ; i++) {
                    while(!Deq.empty() && Deq.back()>X[0][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[0][i][j][k][l]);
                }
                for(i=M ; i<=N ; i++) {
                    if(Deq.front()==X[0][i-M][j][k][l])
                        Deq.pop_front();
                    while(!Deq.empty() && Deq.back()>X[0][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[0][i][j][k][l]);
                    X[1][i-M+1][j][k][l]=min(X[1][i-M+1][j][k][l],Deq.front());
                }
            }
    for(i=1 ; i<=N-M+1 ; i++)
        for(k=1 ; k<=N ; k++)
            for(l=1 ; l<=N ; l++) {
                X[1][i][0][k][l]=INF;
                Deq.clear();
                for(j=1 ; j<=M-1 ; j++) {
                    while(!Deq.empty() && Deq.back()>X[1][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[1][i][j][k][l]);
                }
                for(j=M ; j<=N ; j++) {
                    if(Deq.front()==X[1][i][j-M][k][l])
                        Deq.pop_front();
                    while(!Deq.empty() && Deq.back()>X[1][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[1][i][j][k][l]);
                    X[2][i][j-M+1][k][l]=min(X[2][i][j-M+1][k][l],Deq.front());
                }
            }
    for(i=1 ; i<=N-M+1 ; i++)
        for(j=1 ; j<=N-M+1 ; j++)
            for(l=1 ; l<=N ; l++) {
                X[2][i][j][0][l]=INF;
                Deq.clear();
                for(k=1 ; k<=M-1 ; k++) {
                    while(!Deq.empty() && Deq.back()>X[2][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[2][i][j][k][l]);
                }
                for(k=M ; k<=N ; k++) {
                    if(Deq.front()==X[2][i][j][k-M][l])
                        Deq.pop_front();
                    while(!Deq.empty() && Deq.back()>X[2][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[2][i][j][k][l]);
                    X[3][i][j][k-M+1][l]=min(X[3][i][j][k-M+1][l],Deq.front());
                }
            }
    for(i=1 ; i<=N-M+1 ; i++)
        for(j=1 ; j<=N-M+1 ; j++)
            for(k=1 ; k<=N-M+1 ; k++) {
                X[3][i][j][k][0]=INF;
                Deq.clear();
                for(l=1 ; l<=M-1 ; l++) {
                    while(!Deq.empty() && Deq.back()>X[3][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[3][i][j][k][l]);
                }
                for(l=M ; l<=N ; l++) {
                    if(Deq.front()==X[3][i][j][k][l-M])
                        Deq.pop_front();
                    while(!Deq.empty() && Deq.back()>X[3][i][j][k][l])
                        Deq.pop_back();
                    Deq.push_back(X[3][i][j][k][l]);
                    Y[i][j][k][l-M+1]=min(Y[i][j][k][l-M+1],Deq.front());
                }
            }
}

void output(void) {

    int i, j, k, l;

    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(l=1 ; l<=N-M+1 ; l++)
                    printf("%d ",Y[i][j][k][l]);
}

int main(void) {

    input();
    process();
    output();

    return 0;
}

• Subtask 1: 100 / 100