# | 제출 시각 | 아이디 | 문제 | 언어 | 결과 | 실행 시간 | 메모리 |
---|---|---|---|---|---|---|---|
333659 | apostoldaniel854 | 최솟값 배열 (IZhO11_hyper) | C++14 | 515 ms | 44396 KiB |
이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define pb push_back
#define dbg(x) cerr << #x << " " << x << "\n"
struct MinDeque {
deque <pair <int, int>> dq;
inline void add (int val, int index) {
while (dq.size () && dq.back ().first >= val)
dq.pop_back ();
dq.push_back ({val, index});
}
inline int extract (int bound) {
while (dq.size () && dq.front ().second < bound)
dq.pop_front ();
return dq.front ().first;
}
};
int main () {
ios::sync_with_stdio (false);
cin.tie (0); cout.tie (0);
int n, m;
cin >> n >> m;
vector <vector <vector <vector <int>>>> X, Y;
X.resize (n + 1);
for (int i = 1; i <= n; i++) {
X[i].resize (n + 1);
for (int j = 1; j <= n; j++) {
X[i][j].resize (n + 1);
for (int k = 1; k <= n; k++) {
X[i][j][k].resize (n + 1);
for (int l = 1; l <= n; l++) {
cin >> X[i][j][k][l];
}
}
}
}
Y = X;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int k = 1; k <= n; k++) {
MinDeque dq;
for (int l = 1; l <= n; l++) {
dq.add (Y[i][j][k][l], l);
if (l >= m)
Y[i][j][k][l - m + 1] = dq.extract (l - m + 1);
}
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
for (int l = 1; l <= n - m + 1; l++) {
MinDeque dq;
for (int k = 1; k <= n; k++) {
dq.add (Y[i][j][k][l], k);
if (k >= m)
Y[i][j][k - m + 1][l] = dq.extract (k - m + 1);
}
}
for (int i = 1; i <= n; i++)
for (int k = 1; k <= n - m + 1; k++)
for (int l = 1; l <= n - m + 1; l++) {
MinDeque dq;
for (int j = 1; j <= n; j++) {
dq.add (Y[i][j][k][l], j);
if (j >= m)
Y[i][j - m + 1][k][l] = dq.extract (j - m + 1);
}
}
for (int j = 1; j <= n - m + 1; j++)
for (int k = 1; k <= n - m + 1; k++)
for (int l = 1; l <= n - m + 1; l++) {
MinDeque dq;
for (int i = 1; i <= n; i++) {
dq.add (Y[i][j][k][l], i);
if (i >= m)
Y[i - m + 1][j][k][l] = dq.extract (i - m + 1);
}
}
for (int i = 1; i <= n - m + 1; i++)
for (int j = 1; j <= n - m + 1; j++)
for (int k = 1; k <= n - m + 1; k++)
for (int l = 1; l <= n - m + 1; l++)
cout << Y[i][j][k][l] << " ";
cout << "\n";
return 0;
}
# | Verdict | Execution time | Memory | Grader output |
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Fetching results... |