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>
#pragma GCC optimize ("Ofast", "unroll-loops")
using namespace std;
#define ll long long
#define int ll
#define FOR(i,a,b) for (int i = (a); i<(b); i++)
#define REP(i,n) FOR(i,0,n)
#define REP1(i,n) FOR(i,1,n+1)
#define RREP(i,n) for (int i=(n)-1; i>=0; i--)
#define f first
#define s second
#define pb push_back
#define ALL(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define SQ(x) (x)*(x)
#define pii pair<int, int>
#define pdd pair<double ,double>
#define pcc pair<char, char>
#define endl '\n'
//#define TOAD
#ifdef TOAD
#define bug(x) cerr<<__LINE__<<": "<<#x<<" is "<<x<<endl
#define IOS()
#else
#define bug(...)
#define IOS() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#endif
const ll inf = (1ll<<60);
const int iinf=2147483647;
const ll mod = 1e9+7;
const ll maxn=2005;
const double PI=acos(-1);
ll pw(ll x, ll p, ll m=mod){
ll ret=1;
while (p>0){
if (p&1){
ret*=x;
ret%=m;
}
x*=x;
x%=m;
p>>=1;
}
return ret;
}
ll inv(ll a, ll m=mod){
return pw(a,m-2);
}
int a[2][maxn][maxn];
signed main(){
IOS();
int h,w; cin>>h>>w;
int fmx=0, fmn=inf;
REP(i,h){
REP(j,w){
cin>>a[0][i][j];
fmx=max(fmx, a[0][i][j]);
fmn=min(fmn, a[0][i][j]);
a[1][i][w-j-1]=a[0][i][j];
}
}
/*REP(id, 4){
if (id%2) swap(h, w);
REP(i,h){
REP(j,w){
cout<<a[id][i][j]<<' ';
}
cout<<endl;
}
if (id%2) swap(h, w);
cout<<endl;
}
*/
int ans=inf;
FOR(id, 0, 2){
int l=0, r=fmx-fmn;
while(l<=r){
if (l==r){
ans=min(ans, l);
break;
}
int mid=(l+r)/2;
bool gg=false;
int prevd=h;
REP(i,w){
int k=-1;
while(k+1<h&&a[id][k+1][i]<=fmn+mid){
k++;
}
k=min(k, prevd);
prevd=k;
//cout<<k<<endl;
//cout<<acmx<<' '<<acmn<<endl;
FOR(j, k+1, h){
if (fmx-mid>a[id][j][i]){
gg=1;
break;
}
}
if (gg) break;
}
if (gg){
l=mid+1;
}
else {
r=mid;
}
}
l=0; r=fmx-fmn;
while(l<=r){
if (l==r){
ans=min(ans, l);
break;
}
int mid=(l+r)/2;
bool gg=false;
int prevd=h;
REP(i,w){
int k=-1;
while(k+1<h&&a[id][k+1][i]>=fmx-mid){
k++;
}
k=min(k, prevd);
prevd=k;
//cout<<k<<endl;
//cout<<acmx<<' '<<acmn<<endl;
FOR(j, k+1, h){
if (mid+fmn<a[id][j][i]){
gg=1;
break;
}
}
if (gg) break;
}
if (gg){
l=mid+1;
}
else {
r=mid;
}
}
}
cout<<ans<<endl;
}
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