Submission #13260

#	Submission time	Handle	Problem	Language	Result	Execution time	Memory
13260	2015-02-06T18:03:48 Z	baneling100	Hyper-minimum (IZhO11_hyper)	C++	100 / 100	499 ms	34392 KB

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

100.0 / 100.0

#	Verdict	Execution time	Memory	Grader output
1	Correct	0 ms	34392 KB	Output is correct
2	Correct	0 ms	34392 KB	Output is correct
3	Correct	0 ms	34392 KB	Output is correct
4	Correct	2 ms	34392 KB	Output is correct
5	Correct	3 ms	34392 KB	Output is correct
6	Correct	20 ms	34392 KB	Output is correct
7	Correct	10 ms	34392 KB	Output is correct
8	Correct	16 ms	34392 KB	Output is correct
9	Correct	31 ms	34392 KB	Output is correct
10	Correct	33 ms	34392 KB	Output is correct
11	Correct	77 ms	34392 KB	Output is correct
12	Correct	188 ms	34392 KB	Output is correct
13	Correct	150 ms	34392 KB	Output is correct
14	Correct	267 ms	34392 KB	Output is correct
15	Correct	431 ms	34392 KB	Output is correct
16	Correct	164 ms	34392 KB	Output is correct
17	Correct	342 ms	34392 KB	Output is correct
18	Correct	499 ms	34392 KB	Output is correct
19	Correct	364 ms	34392 KB	Output is correct
20	Correct	284 ms	34392 KB	Output is correct