HDU-2888 Check Corners 二维RMQ

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2888

　　模板题。解题思路如下（转载别人写的)：

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)

//STATUS:C++_AC_4109MS_30160KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef long long LL;
typedef unsigned long long ULL;
//const
const int N=310;
const int INF=0x3f3f3f3f;
const int MOD=1e+7,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End
int val[N][N],dp[N][N][9][9];
int n,m,Q;
void init()
{
    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 RMQ2D(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 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()
{
 //   freopen("in.txt","r",stdin);
    int i,j,ans;
    int x1,y1,x2,y2;
    while(~scanf("%d%d",&n,&m))
    {
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                scanf("%d",&val[i][j]);
        init();
        scanf("%d",&Q);
        while(Q--){
            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
            ans=RMQ2D(x1,y1,x2,y2);
            if(ans==val[x1][y1] || ans==val[x2][y2]
               || ans==val[x1][y2] || ans==val[x2][y1]){
                printf("%d yes
",ans);
            }
            else printf("%d no
",ans);
        }
    }
    return 0;
}