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=;
 const int INF=0x3f3f3f3f;
 const int MOD=1e+,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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][][];
 int n,m,Q;
 void init()
 {
     for(int row = ; row <= n; row++)
         for(int col = ; col <=m; col++)
             dp[row][col][][] = val[row][col];
     int mx = log(double(n)) / log(2.0);
     int my = log(double(m)) / log(2.0);
     for(int i=; i<= mx; i++)
     {
         for(int j = ; j<=my; j++)
         {
             if(i ==  && j ==) continue;
             for(int row = ; row+(<<i)- <= n; row++)
             {
                 for(int col = ; col+(<<j)- <= m; col++)
                 {
                     if(i == )//y轴二分
                         dp[row][col][i][j]=max(dp[row][col][i][j-],dp[row][col+(<<(j-))][i][j-]);
                     else//x轴二分
                         dp[row][col][i][j]=max(dp[row][col][i-][j],dp[row+(<<(i-))][col][i-][j]);
                 }
             }
         }
     }
 }
 int RMQ2D(int x1,int y1,int x2,int y2)
 {
     int kx = log(double(x2-x1+)) / log(2.0);
     int ky = log(double(y2-y1+)) / log(2.0);
     int m1 = dp[x1][y1][kx][ky];
     int m2 = dp[x2-(<<kx)+][y1][kx][ky];
     int m3 = dp[x1][y2-(<<ky)+][kx][ky];
     int m4 = dp[x2-(<<kx)+][y2-(<<ky)+][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=;i<=n;i++)
             for(j=;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\n",ans);
             }
             else printf("%d no\n",ans);
         }
     }
     return ;
 }