[bzoj1047][HAOI2007]理想的正方形【单调队列】

【题目链接】
　　http://www.lydsy.com/JudgeOnline/problem.php?id=1047
【题解】
　　横着做一遍单调队列得出每个点往右n$n$个的最大/最小值。
　　接下来用之前的结果竖着做一遍单调队列得出每个n∗n$n\ast n$矩形的最大/最小值。
　　复杂度O(n2)$O(n^2)$

/* --------------
    user Vanisher
    problem bzoj-1047 
----------------*/
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    N       1010
using namespace std;
int read(){
    int tmp=0, fh=1; char ch=getchar();
    while (ch<'0'||ch>'9'){if (ch=='-') fh=-1; ch=getchar();}
    while (ch>='0'&&ch<='9'){tmp=tmp*10+ch-'0'; ch=getchar();}
    return tmp*fh;
}
int n,m,k,mp[N][N],q[N],mx[N][N],mn[N][N],fx[N][N],fn[N][N];
int main(){
    n=read(), m=read(), k=read();
    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            mp[i][j]=read();
    for (int i=1; i<=n; i++){
        int pl=1, pr=0;
        for (int j=1; j<=k-1; j++){
            while (pl<=pr&&mp[i][j]>mp[i][q[pr]]) pr--;
            q[++pr]=j;
        }
        for (int j=k; j<=m; j++){
            while (q[pl]<=j-k) pl++;
            while (pl<=pr&&mp[i][j]>mp[i][q[pr]]) pr--;
            q[++pr]=j;
            mx[i][j-k+1]=mp[i][q[pl]];
        }
    }
    for (int i=1; i<=n; i++){
        int pl=1, pr=0;
        for (int j=1; j<=k-1; j++){
            while (pl<=pr&&mp[i][j]<mp[i][q[pr]]) pr--;
            q[++pr]=j;
        }
        for (int j=k; j<=m; j++){
            while (q[pl]<=j-k) pl++;
            while (pl<=pr&&mp[i][j]<mp[i][q[pr]]) pr--;
            q[++pr]=j;
            mn[i][j-k+1]=mp[i][q[pl]];
        }
    }
    for (int i=1; i<=m-k+1; i++){
        int pl=1, pr=0;
        for (int j=1; j<=k-1; j++){
            while (pl<=pr&&mx[j][i]>mx[q[pr]][i]) pr--;
            q[++pr]=j;
        }
        for (int j=k; j<=n; j++){
            while (q[pl]<=j-k) pl++;
            while (pl<=pr&&mx[j][i]>mx[q[pr]][i]) pr--;
            q[++pr]=j;
            fx[j-k+1][i]=mx[q[pl]][i];
        }
    }
    for (int i=1; i<=m-k+1; i++){
        int pl=1, pr=0;
        for (int j=1; j<=k-1; j++){
            while (pl<=pr&&mn[j][i]<mn[q[pr]][i]) pr--;
            q[++pr]=j;
        }
        for (int j=k; j<=n; j++){
            while (q[pl]<=j-k) pl++;
            while (pl<=pr&&mn[j][i]<mn[q[pr]][i]) pr--;
            q[++pr]=j;
            fn[j-k+1][i]=mn[q[pl]][i];
        }
    }
    int ans=inf;
    for (int i=1; i<=n-k+1; i++)
        for (int j=1; j<=m-k+1; j++)
            ans=min(ans,fx[i][j]-fn[i][j]);
    printf("%d
",ans);
    return 0;   
}