理想的正方形

看了一篇题解!来练习单调队列优化的DP

用单调队列分别维护行与列

在维护的同事随便维护最大值和最小值

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 1010
using namespace std ;
int n , m , k , front , Front , back ,Back , ans ;
int a[maxn][maxn] ,q[maxn] , Q[maxn] ;
int x[maxn][maxn] , X[maxn][maxn] ;
int y[maxn][maxn] , Y[maxn][maxn] ;

int main () {
	cin >> n >> m >> k ;
	for(int i = 1 ; i <= n ; i ++) {
		for(int j  = 1 ; j <= m ; j ++) {
			cin >> a[i][j] ;
		}
	}
	for(int i = 1 ; i <= n ; i  ++) {
		Front = Back = front = back = Q[1] = q[1] = 1 ;
		for(int j = 2 ; j <= m ; j ++) {
			while(a[i][j] >= a[i][Q[Back]] && Front <= Back) Back -- ;
			while(a[i][j] <= a[i][q[back]] && front <= back) back -- ;
			Back ++ ;back ++ ;
			Q[Back] = j ;q[back] = j ;
			while(j-Q[Front] >= k) Front ++ ;
			while(j-q[front] >= k) front ++ ;
			if(j >= k) X[i][j-k+1] = a[i][Q[Front]] ,
			           x[i][j-k+1] = a[i][q[front]] ;
 		}
	}
	/*for(int i = 1 ; i <= m-k+1 ; i ++) {
		Front = Back = front = back = Q[1] = q[1] = 1 ;
		for(int j = 2 ; j <= n ; j ++) {
			while(X[j][i] >= X[Q[Back]][i] && Front <= Back) Back -- ;
			while(x[j][i] <= x[q[back]][i] && front <= back) back -- ;
			Back ++ , back ++ ;
			Q[Back] = j , q[back] = j ;
			while(j-Q[Front] >= k) Front ++ ;
			while(j-q[front] >= k) front ++ ;
			if(j >= k) Y[j-k+1][i] = X[Q[Front]][i] , y[i-k+1][i] = x[q[front]][i] ;
		}
	}*/
	for (int I=1;I<=m-k+1;I++)
        {
            Front=Back=front=back=Q[1]=q[1]=1;
            for (int i=2;i<=n;i++)
                {
                    while (X[i][I]>=X[Q[Back]][I]&&Front<=Back) Back--;
                    while (x[i][I]<=x[q[back]][I]&&front<=back) back--;
                    Back++;back++;Q[Back]=i;q[back]=i;
                    while (i-Q[Front]>=k) Front++;
                    while (i-q[front]>=k) front++;
                    if (i>=k) Y[i-k+1][I]=X[Q[Front]][I],y[i-k+1][I]=x[q[front]][I];
                }
        }
	ans = 0x3f3f3f3f ;
	for(int i = 1 ; i <= n-k+1 ; i ++) {
		for(int j = 1 ; j <= m-k+1 ; j ++) {
			ans = min(ans,Y[i][j]-y[i][j]) ;
		}
	}
	printf("%d
",ans); 
	return 0 ;
}