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 ;
 }