HDU2888 Check Corners（二维RMQ）

有一个矩阵，每次查询一个子矩阵，判断这个子矩阵的最大值是不是在这个子矩阵的四个角上

裸的二维RMQ

 #pragma comment(linker, "/STACK:1677721600")

 #include <map>

 #include <set>

 #include <stack>

 #include <queue>

 #include <cmath>

 #include <ctime>

 #include <vector>

 #include <cstdio>

 #include <cctype>

 #include <cstring>

 #include <cstdlib>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 #define INF 0x3f3f3f3f

 #define inf (-((LL)1<<40))

 #define lson k<<1, L, (L + R)>>1

 #define rson k<<1|1,  ((L + R)>>1) + 1, R

 #define mem0(a) memset(a,0,sizeof(a))

 #define mem1(a) memset(a,-1,sizeof(a))

 #define mem(a, b) memset(a, b, sizeof(a))

 #define FIN freopen("in.txt", "r", stdin)

 #define FOUT freopen("out.txt", "w", stdout)

 #define rep(i, a, b) for(int i = a; i <= b; i ++)

 #define dec(i, a, b) for(int i = a; i >= b; i --)

 template<class T> T MAX(T a, T b) { return a > b ? a : b; }

 template<class T> T MIN(T a, T b) { return a < b ? a : b; }

 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;    }

 //typedef __int64 LL;

 typedef long long LL;

 const int MAXN =  + ;

 const int MAXM = ;

 const double eps = 1e-;

 LL MOD = ;

 int m, n;

 int mx[][][][];

 int idx[], q, lx, ly, rx, ry;

 void rmq_init(int m, int n) {

     for(int i = ; (<<i) <= m; i ++) {

         for(int j = ; (<<j) <= n; j ++) {

             if(i ==  && j == ) continue;

             int len2 = ( << j), len1 = ( << i);

             for(int x = ; x + len1 -  <= m; x ++) {

                 for(int y = ; y + len2 -  <= n; y ++) {

                     if(i == ) mx[x][i][y][j] = max(mx[x][i][y][j - ], mx[x][i][y + (len2 >> )][j - ]);

                     else mx[x][i][y][j] = max(mx[x][i - ][y][j], mx[x + (len1 >> )][i - ][y][j]);

                 }

             }

         }

     }

     for(int i = ; i <= m || i <= n; i ++) {

         idx[i] = ;

         while(( << (idx[i] + )) <= i) idx[i] ++;

     }

 }

 int rmq(int lx, int rx, int ly, int ry) {

     int a = idx[rx - lx + ], la = ( << a);

     int b = idx[ry - ly + ], lb = ( << b);

     return max(max(max(mx[lx][a][ly][b],

                        mx[rx - la + ][a][ly][b]),

                        mx[lx][a][ry - lb + ][b]),

                        mx[rx - la + ][a][ry - lb + ][b]);

 }

 int main()

 {

     //FIN;

     while(~scanf("%d %d", &m, &n)) {

         mem0(mx);

         rep (i, , m) rep (j, , n)

             scanf("%d", &mx[i][][j][]);

         rmq_init(m, n);

         scanf("%d", &q);

         while(q--) {

             scanf("%d %d %d %d", &lx, &ly, &rx, &ry);

             int ma = rmq(lx, rx, ly, ry);

             printf("%d %s\n", ma, ma == mx[lx][][ly][] || ma == mx[lx][][ry][]

                                || ma == mx[rx][][ly][] || ma == mx[rx][][ry][]

                                ? "yes" : "no");

         }

     }

     return ;

 }