HDU2888二维RMQ dp[row][col][i][j] 表示[row,row+2^i-1]x[col,col+2^j-1] 二维区间内的最小值 这是RMQ-ST算法的核心: 倍增思想 == min( [row,row+ 2^(i-1)-1]x[col,col+2^j-1], [row+2^(i-1),row+2^i-1]x[col,col+2^j-1] ) = min(dp[row][col][i-1][j], dp[row+(1<<(i-1))][col][i-1][j] ) //y轴不变,x轴二分　（i!=0) 或 == min( [row,row+2^i-1]x[col,col+2^(j-1)-1], [row,row+2^i-1]x[col+2^(i-1),col+2^j-1] ) = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] ) //x轴不变,y轴二分　(j!=0) 即: dp[row][col][i][j] = min(dp[row][col][i-1][j], dp[row + (1<<(i-1))][col][i-1][j] )
或 = min(dp[row][col][i][j-1], dp[row][col+(1<<(j-1))][i][j-1] ) 查询[x1,x2]x[y1,y2] 令 kx = (int)log2(x2-x1+1); ky = (int)log2(y2-y1+1); 查询结果为 m1 = dp[x1][y1][kx][ky] = dp[x1][y1][kx][ky]; m2 = dp[x2-2^kx+1][y1][kx]ky] = dp[x2-(1<<kx)+1][y1][kx][ky]; m3 = dp[x1][y2-2^ky+1][kx][ky] = dp[x1][y2-(1<<ky)+1][kx][ky]; m4 = dp[x2-2^kx+1][y2-2^ky+1][kx][ky] = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky]; 结果 = min(m1,m2,m3,m4)

#include<iostream>
#include<cmath>
using namespace std;
const int maxn = 301;
int val[maxn][maxn];

int M,N;

//RMQ 2D
int dp[maxn][maxn][9][9];
void RMQ_2D_PRE()
{
    for(int row = 1; row <= N; row++)
        for(int col = 1; col <=M; col++)
            dp[row][col][0][0] = val[row][col];
    int mx = log(double(N)) / log(2.0);
    int my = log(double(M)) / log(2.0);
    for(int i=0; i<= mx; i++)
    {
        for(int j = 0; j<=my; 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 <= M; col++)
                {
                    if(i == 0)//y轴二分
                        dp[row][col][i][j]=max(dp[row][col][i][j-1],dp[row][col+(1<<(j-1))][i][j-1]);  
                    else//x轴二分
                        dp[row][col][i][j]=max(dp[row][col][i-1][j],dp[row+(1<<(i-1))][col][i-1][j]);  
                }
            }
        }
    }
}
int RMQ_2D(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 = dp[x1][y1][kx][ky];
    int m2 = dp[x2-(1<<kx)+1][y1][kx][ky];
    int m3 = dp[x1][y2-(1<<ky)+1][kx][ky];
    int m4 = dp[x2-(1<<kx)+1][y2-(1<<ky)+1][kx][ky];
    return max( max(m1,m2) , max(m3,m4));
}
//END

int  main()
{
    int Q,x1,y1,x2,y2,i,j;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        for(i = 1; i <= N; i++)
            for(j = 1; j <= M; j++)
                scanf("%d",&val[i][j]);
        RMQ_2D_PRE();
        scanf("%d",&Q);
        while(Q--)
        {
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            int ans = RMQ_2D(x1,x2,y1,y2);
            printf("%d",ans);
            if(ans == val[x1][y1] || ans == val[x2][y1] || ans == val[x1][y2] || ans == val[x2][y2])
                printf(" yes\n");
            else printf(" no\n");
        }
    }
    return 0;
}