[luoguP2216] [HAOI2007]理想的正方形（二维单调队列）

传送门

1.先弄个单调队列求出每一行的区间为n的最大值最小值。

2.然后再搞个单调队列求1所求出的结果的区间为n的最大值最小值

3.最后扫一遍就行

懒得画图，自己体会吧。

——代码

#include <cstdio>
#include <iostream>

using namespace std;

const int MAXN = 1001;
int a, b, n, h, t;
long long c[MAXN][MAXN], q[MAXN], max1[MAXN][MAXN], min1[MAXN][MAXN], max2[MAXN][MAXN], min2[MAXN][MAXN], ans = 1000000001;

inline void work1(int k)
{
    int i, j;
    h = 1, t = 0;
    for(i = 1; i <= b; i++)
    {
        while(h <= t && c[k][q[t]] > c[k][i]) t--;
        q[++t] = i;
        while(h <= t && q[h] <= i - n) h++; 
        min1[k][i] = c[k][q[h]];
    }
    h = 1; t = 0;
    for(i = 1; i <= b; i++)
    {
        while(h <= t && c[k][q[t]] < c[k][i]) t--;
        q[++t] = i;
        while(h <= t && q[h] <= i - n) h++; 
        max1[k][i] = c[k][q[h]];
    }
}

inline void work2(int k)
{
    int i, j;
    h = 1, t = 0;
    for(i = 1; i <= a; i++)
    {
        while(h <= t && min1[q[t]][k] > min1[i][k]) t--;
        q[++t] = i;
        while(h <= t && q[h] <= i - n) h++; 
        min2[i][k] = min1[q[h]][k];
    }
    h = 1; t = 0;
    for(i = 1; i <= a; i++)
    {
        while(h <= t && max1[q[t]][k] < max1[i][k]) t--;
        q[++t] = i;
        while(h <= t && q[h] <= i - n) h++; 
        max2[i][k] = max1[q[h]][k];
    }
}

int main()
{
    int i, j;
    scanf("%d %d %d", &a, &b, &n);
    for(i = 1; i <= a; i++)
        for(j = 1; j <= b; j++)
            scanf("%lld", &c[i][j]);
    for(i = 1; i <= a; i++) work1(i);
    for(i = 1; i <= b; i++) work2(i);
    for(i = n; i <= a; i++)
        for(j = n; j <= b; j++)
            ans = min(ans, max2[i][j] - min2[i][j]);
    printf("%lld", ans);
    return 0;
}