题目来源：POJ 2019 Cornfields

题意：求正方形二维区间最大最小值的差

思路：直接二维ST搞 试模版而已

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 255;
int dp[maxn][maxn][8][8];
int dp2[maxn][maxn][8][8];
int a[maxn][maxn];
int n, m;
void RMQ_init(int n)
{
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
		{
			dp[i][j][0][0] = a[i][j];
			dp2[i][j][0][0] = a[i][j];
		}
	int k = (int) (log((double)n + 0.2) / log(2.0));
	for(int x = 0; x <= k; x++)
		for(int y = 0; y <= k; y++)
		{
			if(!x && !y)
				continue;
			for(int i = 0; i + (1<<x) - 1 <= n; i++)
				for(int j = 0; j + (1<<y) - 1 <= n; j++)
				{
					if(x == 0)
					{
						dp[i][j][x][y] = max(dp[i][j][x][y-1], dp[i][j+(1<<(y-1))][x][y-1]);
						dp2[i][j][x][y] = min(dp2[i][j][x][y-1], dp2[i][j+(1<<(y-1))][x][y-1]);
					}
					else
					{
						dp[i][j][x][y] = max(dp[i][j][x-1][y], dp[i+(1<<(x-1))][j][x-1][y]);
						dp2[i][j][x][y] = min(dp2[i][j][x-1][y], dp2[i+(1<<(x-1))][j][x-1][y]);
					}
				}
		}
}
int RMQ(int x1, int y1, int x2, int y2)
{
	int kx = log(double(x2 - x1 + 1)) / log(2.0);
	int ky = log(double(y2 - y1 + 1)) / log(2.0);
	int ans1 = 0;
	ans1 = max(ans1, dp[x1][y1][kx][ky]);
	ans1 = max(ans1, dp[x2-(1<<kx)+1][y1][kx][ky]);
	ans1 = max(ans1, dp[x1][y2-(1<<ky)+1][kx][ky]);
	ans1 = max(ans1, dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]);
	
	int ans2 = 999999999;
	ans2 = min(ans2, dp2[x1][y1][kx][ky]);
	ans2 = min(ans2, dp2[x2-(1<<kx)+1][y1][kx][ky]);
	ans2 = min(ans2, dp2[x1][y2-(1<<ky)+1][kx][ky]);
	ans2 = min(ans2, dp2[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]);
	
	return ans1 - ans2;
}

int main()
{
	int q;
	scanf("%d %d %d", &n, &m, &q);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= n; j++)
			scanf("%d", &a[i][j]);
	RMQ_init(n);
	//printf("Case %d:\n", cas++);
	while(q--)
	{
		int l, r;
		scanf("%d %d", &l, &r);
		int ll = l + m - 1;
		int rr = r + m - 1;
		if(ll > n)
			ll = n;
		if(rr > n)
			rr = n;
		printf("%d\n", RMQ(l, r, ll, rr));
	}
	return 0;
}