poj--2019 Cornfields 2维RMQ

题目大意： 给定一个 N*N 的数组   求以（x1， y1） 为左上角  （x1 + b -1 ，y1 + b -1)为右下角 这个b*b的范围内最大值减最小值

看到最大值最小值当然想到RMQ啦

//Accepted    27392K    594MS    C++    2416B
#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
const int maxn = 301;
int N;
int val[maxn][maxn];
int st_min[maxn][maxn][9][9];
int st_max[maxn][maxn][9][9];
inline int minn(int a,int b){ return a>b?b:a; }
inline int maxx(int a,int b){ return a>b?a:b; }
void RMQ_2D_PRE()
{
    for(int row = 1; row <= N; row++)
        for(int col = 1; col<=N; col++)
        {
            st_min[row][col][0][0] = val[row][col];
            st_max[row][col][0][0] = val[row][col];
        }
        
        int m = log(double(N)) / log(2.0);
        
        for(int i=0; i<=m; i++)
            for(int j=0; j<=m; j++)
            {
                if(i == 0 && j==0)
                    continue;
                
                for(int row = 1; row+(1<<i)-1 <= N; row++)
                    for(int col = 1; col+(1<<j)-1 <= N; col++)
                    {
                        if(i == 0)
                        {
                            st_min[row][col][i][j] = minn(st_min[row][col][i][j-1] , st_min[row][col+(1<<(j-1))][i][j-1]);  //水平划分
                            st_max[row][col][i][j] = maxx(st_max[row][col][i][j-1] , st_max[row][col+(1<<(j-1))][i][j-1]);  //水平划分
                        }
                        else
                        {
                            st_min[row][col][i][j] = minn(st_min[row][col][i-1][j] , st_min[row+(1<<(i-1))][col][i-1][j]);    //竖直划分
                            st_max[row][col][i][j] = maxx(st_max[row][col][i-1][j] , st_max[row+(1<<(i-1))][col][i-1][j]);    //竖直划分
                        }
                    }
            }
}
int RMQ_2D_min(int x1,int x2,int y1,int y2)
{
    int kx = log(double(x2 - x1 +1)) / log(2.0);
    
    int ky = log(double(y2 - y1 +1)) / log(2.0);
    
    int m1 = st_min[x1][y1][kx][ky];
    
    int m2 = st_min[x2-(1<<kx)+1][y1][kx][ky];
    
    int m3 = st_min[x1][y2-(1<<ky)+1][kx][ky];
    
    int m4 = st_min[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
    
    return minn( minn(m1,m2), minn(m3,m4) );

}
int RMQ_2D_max(int x1,int x2,int y1,int y2)
{
    int kx = log(double(x2 - x1 +1)) / log(2.0);
    
    int ky = log(double(y2 - y1 +1)) / log(2.0);
    
    int m1 = st_max[x1][y1][kx][ky];
    
    int m2 = st_max[x2-(1<<kx)+1][y1][kx][ky];
    
    int m3 = st_max[x1][y2-(1<<ky)+1][kx][ky];
    
    int m4 = st_max[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
    
    return maxx( maxx(m1,m2), maxx(m3,m4) );
    
}
int main(void)
{
    
    int M,B;
    int x1,y1;
    while(scanf("%d%d%d",&N,&B,&M)!=EOF)
    {    
        for(int i=1; i<=N; i++)
            for(int j=1; j<=N; j++)
                scanf("%d",&val[i][j]);            
            RMQ_2D_PRE();                    
            while(M--)
            {
                scanf("%d%d",&x1,&y1);            
                printf("%d\n",RMQ_2D_max(x1,x1+B-1,y1,y1+B-1)-RMQ_2D_min(x1,x1+B-1,y1,y1+B-1));
            }
            
    }
    return 0;
}