[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 = ;

 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 = ;

 inline void work1(int k)

 {

     int i, j;

     h = , t = ;

     for(i = ; 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 = ; t = ;

     for(i = ; 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 = , t = ;

     for(i = ; 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 = ; t = ;

     for(i = ; 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 = ; i <= a; i++)

         for(j = ; j <= b; j++)

             scanf("%lld", &c[i][j]);

     for(i = ; i <= a; i++) work1(i);

     for(i = ; 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 ;

 }