我是题面

题意挺清晰的，做法也挺简单的

用单调队列维护以\((i,j)\)为左上角的正方形里最大最小分别是多少，存到数组里，然后遍历找答案，就这样

下面放代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cctype>
#include<queue>
#define ll long long
#define gc getchar
#define maxn 1005
using namespace std;

inline ll read(){
    ll a=0;int f=0;char p=gc();
    while(!isdigit(p)){f|=p=='-';p=gc();}
    while(isdigit(p)){a=(a<<3)+(a<<1)+(p^48);p=gc();}
    return f?-a:a;
}int n,m,l,ans=2147483647,a[maxn][maxn];
int a1[maxn][maxn],a2[maxn][maxn],b1[maxn][maxn],b2[maxn][maxn];

struct ahaha{
    int s,id;
}s1[maxn],s2[maxn];int h1=1,h2=1,t1,t2;

int main(){
    n=read();m=read();l=read();
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            a[i][j]=read();
    for(int i=1;i<=n;++i){
        for(int j=1;j<l;++j){
            while(h1<=t1&&a[i][j]>=s1[t1].s)--t1;
            while(h2<=t2&&a[i][j]<=s2[t2].s)--t2;
            s1[++t1]=s2[++t2]={a[i][j],j};
        }
        for(int j=l;j<=m;++j){
            while(h1<=t1&&j-s1[h1].id>=l)++h1;
            while(h2<=t2&&j-s2[h2].id>=l)++h2;
            while(h1<=t1&&a[i][j]>=s1[t1].s)--t1;
            while(h2<=t2&&a[i][j]<=s2[t2].s)--t2;
            s1[++t1]=s2[++t2]={a[i][j],j};
            a1[i][j-l+1]=s1[h1].s,a2[i][j-l+1]=s2[h2].s;
        }
        h1=h2=1,t1=t2=0;
    }m-=l-1;
    for(int i=1;i<=m;++i){
        for(int j=1;j<l;++j){
            while(h1<=t1&&a1[j][i]>=s1[t1].s)--t1;
            while(h2<=t2&&a2[j][i]<=s2[t2].s)--t2;
            s1[++t1]={a1[j][i],j},s2[++t2]={a2[j][i],j};
        }
        for(int j=l;j<=n;++j){
            while(h1<=t1&&j-s1[h1].id>=l)++h1;
            while(h2<=t2&&j-s2[h2].id>=l)++h2;
            while(h1<=t1&&a1[j][i]>=s1[t1].s)--t1;
            while(h2<=t2&&a2[j][i]<=s2[t2].s)--t2;
            s1[++t1]={a1[j][i],j};s2[++t2]={a2[j][i],j};
            b1[j-l+1][i]=s1[h1].s,b2[j-l+1][i]=s2[h2].s;
        }
        h1=h2=1,t1=t2=0;
    }n-=l-1;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            ans=min(ans,b1[i][j]-b2[i][j]);
    printf("%d\n",ans);
    return 0;
}