[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 ;
 }