Idea: Find the minimum and maximum values of a sliding window of length k for each row, then find the column again, and start traversing at k*k.
#pragma GCC optimize(2)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<stack>
#include<iomanip>
#include<cstring>
#include<time.h>
using namespace std;
typedef long long ll;
#define SIS std::ios::sync_with_stdio(false)
#define space putchar(' ')
#define enter putchar('\n')
#define lson root<<1
#define rson root<<1|1
typedef pair<int,int> PII;
const int mod=1e9+7;
const int N=2e6+10;
const int M=1e5+10;
const int inf=0x3f3f3f3f;
const int maxx=2e5+7;
const double eps=1e-6;
int gcd(int a,int b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
ll qsm(int a,int b,int p)
{
ll res=1%p;
while(b)
{
if(b&1)
res=res*a%p;
a=1ll*a*a%p;
b>>=1;
}
return res;
}
int n, m,k;
int dp[N];
int q[N];
int w[1005][1005];
int rmax[1005][1005],rmin[1005][1005];
void getmin(int a[],int b[],int len)
{
int hh=0,tt=-1;
for(int i=1;i<=len;i++)
{
if(hh<=tt&&q[hh]<=i-k)hh++;
while(hh<=tt&&a[q[tt]]>=a[i])tt--;
q[++tt]=i;
b[i]=a[q[hh]];
}
}
void getmax(int a[],int b[],int len)
{
int hh=0,tt=-1;
for(int i=1;i<=len;i++)
{
if(hh<=tt&&q[hh]<=i-k)hh++;
while(hh<=tt&&a[q[tt]]<=a[i])tt--;
q[++tt]=i;
b[i]=a[q[hh]];
}
}
int main()
{
scanf("%d%d%d",&n,&m,&k);
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
scanf("%d",&w[i][j]);
}
}
int a[1005],b[1005],c[1005];
for(int i=1;i<=n;i++){
getmin(w[i],rmin[i],m);
getmax(w[i],rmax[i],m);
}
int res=inf;
for(int i=k;i<=m;i++)
{
for(int j=1;j<=n;j++)
a[j]=rmin[j][i];
getmin(a,b,n);
for(int j=1;j<=n;j++)
a[j]=rmax[j][i];
getmax(a,c,n);
for(int j=k;j<=n;j++) res=min(res,c[j]-b[j]);
}
printf("%d",res);
return 0;
}