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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;
int H,W,g[2000][2000];
int l[2][2000], r[2][2000];
pi lo = {0,0}, hi = {0,0};
int L[2000], R[2000];
bool leftIncrease(int x) {
F0R(i,H) {
R[i] = r[x][i];
if (i) R[i] = max(R[i],R[i-1]);
}
F0Rd(i,H) {
L[i] = l[x^1][i];
if (i < H-1) L[i] = min(L[i],L[i+1]);
}
F0R(i,H) if (R[i] >= L[i]) return 0;
return 1;
}
bool leftDecrease(int x) {
F0Rd(i,H) {
R[i] = r[x][i];
if (i < H-1) R[i] = max(R[i],R[i+1]);
}
F0R(i,H) {
L[i] = l[x^1][i];
if (i) L[i] = min(L[i],L[i-1]);
}
F0R(i,H) if (R[i] >= L[i]) return 0;
return 1;
}
bool ok(int mid) {
F0R(i,H) l[0][i] = l[1][i] = MOD, r[0][i] = r[1][i] = -MOD;
F0R(i,H) F0R(j,W) {
if (g[hi.f][hi.s]-g[i][j] > mid) r[0][i] = max(r[0][i],j), l[0][i] = min(l[0][i],j);
if (g[i][j]-g[lo.f][lo.s] > mid) r[1][i] = max(r[1][i],j), l[1][i] = min(l[1][i],j);
}
return leftIncrease(0) || leftDecrease(0) || leftIncrease(1) || leftDecrease(1);
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> H >> W;
F0R(i,H) F0R(j,W) {
cin >> g[i][j];
if (g[i][j] < g[lo.f][lo.s]) lo = {i,j};
if (g[i][j] > g[hi.f][hi.s]) hi = {i,j};
}
int lo = 0, hi = 1000000000;
while (lo < hi) {
int mid = (lo+hi)/2;
if (ok(mid)) hi = mid;
else lo = mid+1;
}
cout << lo;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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