链接：http://poj.org/problem?id=2019

题意：给你一个n*n的矩阵，q次询问，每次询问给出左上角的坐标，询问以这个点为左上角的b*b的子矩阵中最大值和最小值的差。

思路：二维RMQ的基本应用，网上找的模板

这道题是USACO的，数据很水，暴力也能过。

#include<cstring>
#include<string>
#include<fstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cctype>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<stack>
#include<ctime>
#include<cstdlib>
#include<functional>
#include<cmath>
using namespace std;
#define PI acos(-1.0)
#define MAXN 260
#define eps 1e-7
#define INF 0x3F3F3F3F      //0x7FFFFFFF
#define LLINF 0x7FFFFFFFFFFFFFFF
#define seed 1313131
#define MOD 1000000007
#define ll long long
#define ull unsigned ll
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int a[MAXN][MAXN];
int maxm[MAXN][MAXN][11],minm[MAXN][MAXN][11];
void RMQ(int num){
    int i,j,k;
    for(i = 1; i <= num; i++){
        for(j = 1; j <= num; j++){
            maxm[i][j][0] = minm[i][j][0] = a[i][j];
        }
    }
    for(i = 1; i <= num; i++){
        for(k = 1; (1 << k) <= num; k++){
            for(j = 1; j + (1 << k) - 1 <= num; j++){
                maxm[i][j][k] = max(maxm[i][j][k-1],maxm[i][j+(1<<(k-1))][k-1]);
                minm[i][j][k] = min(minm[i][j][k-1],minm[i][j+(1<<(k-1))][k-1]);
            }
        }
    }
}
int main(){
    int i,j,n,b,q;
    int top,left;
    scanf("%d%d%d",&n,&b,&q);
    for(i = 1; i <= n; i++){
        for(j = 1; j <= n; j++){
            scanf("%d", &a[i][j]);
        }
    }
    RMQ(n);
    while(q--){
        scanf("%d%d", &top, &left);
        int k = (int)(log(double(b))/log(2.0));
        int minans = INF, maxans = 0;
        int x = left, y = left + b - 1;
        for(i = top; i < top + b; i++){
            maxans = max(maxans,max(maxm[i][x][k],maxm[i][y-(1<<k)+1][k]));
            minans = min(minans,min(minm[i][x][k],minm[i][y-(1<<k)+1][k]));
        }
        printf("%d\n",maxans - minans);
    }
    return 0;
}

POJ--2019--Cornfields【二维RMQ】