# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
572246 | SSRS | 최솟값 배열 (IZhO11_hyper) | C++14 | 790 ms | 22320 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <bits/stdc++.h>
using namespace std;
vector<int> slide_min(vector<int> &A, int M){
int N = A.size();
deque<int> dq;
vector<int> ans(N - M + 1);
for (int i = 0; i < N; i++){
while (!dq.empty()){
if (A[dq.back()] >= A[i]){
dq.pop_back();
} else {
break;
}
}
dq.push_back(i);
if (i >= M - 1){
ans[i - M + 1] = A[dq.front()];
if (dq.front() == i - M + 1){
dq.pop_front();
}
}
}
return ans;
}
int main(){
int N, M;
cin >> N >> M;
vector<vector<vector<vector<int>>>> X(N, vector<vector<vector<int>>>(N, vector<vector<int>>(N, vector<int>(N))));
for (int i = 0; i < N; i++){
for (int j = 0; j < N; j++){
for (int k = 0; k < N; k++){
for (int l = 0; l < N; l++){
cin >> X[i][j][k][l];
}
}
}
}
for (int i = 0; i < N; i++){
for (int j = 0; j < N; j++){
for (int k = 0; k < N; k++){
vector<int> A = X[i][j][k];
vector<int> B = slide_min(A, M);
for (int l = 0; l <= N - M; l++){
X[i][j][k][l] = B[l];
}
}
}
}
for (int i = 0; i < N; i++){
for (int j = 0; j < N; j++){
for (int k = 0; k <= N - M; k++){
vector<int> A(N);
for (int l = 0; l < N; l++){
A[l] = X[i][j][l][k];
}
vector<int> B = slide_min(A, M);
for (int l = 0; l <= N - M; l++){
X[i][j][l][k] = B[l];
}
}
}
}
for (int i = 0; i < N; i++){
for (int j = 0; j <= N - M; j++){
for (int k = 0; k <= N - M; k++){
vector<int> A(N);
for (int l = 0; l < N; l++){
A[l] = X[i][l][j][k];
}
vector<int> B = slide_min(A, M);
for (int l = 0; l <= N - M; l++){
X[i][l][j][k] = B[l];
}
}
}
}
for (int i = 0; i <= N - M; i++){
for (int j = 0; j <= N - M; j++){
for (int k = 0; k <= N - M; k++){
vector<int> A(N);
for (int l = 0; l < N; l++){
A[l] = X[l][i][j][k];
}
vector<int> B = slide_min(A, M);
for (int l = 0; l <= N - M; l++){
X[l][i][j][k] = B[l];
}
}
}
}
for (int i = 0; i <= N - M; i++){
for (int j = 0; j <= N - M; j++){
for (int k = 0; k <= N - M; k++){
for (int l = 0; l <= N - M; l++){
cout << X[i][j][k][l];
if (i < N - M || j < N - M || k < N - M || l < N - M){
cout << ' ';
}
}
}
}
}
cout << endl;
}
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |