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 = 50000 + 100;
const int MAXM = 110000;
const double eps = 1e-8;
LL MOD = 1000000007;

int m, n;
int mx[302][9][302][9];
int idx[302], q, lx, ly, rx, ry;

void rmq_init(int m, int n) {
    for(int i = 0; (1<<i) <= m; i ++) {
        for(int j = 0; (1<<j) <= n; j ++) {
            if(i == 0 && j == 0) continue;
            int len2 = (1 << j), len1 = (1 << i);
            for(int x = 1; x + len1 - 1 <= m; x ++) {
                for(int y = 1; y + len2 - 1 <= n; y ++) {
                    if(i == 0) mx[x][i][y][j] = max(mx[x][i][y][j - 1], mx[x][i][y + (len2 >> 1)][j - 1]);
                    else mx[x][i][y][j] = max(mx[x][i - 1][y][j], mx[x + (len1 >> 1)][i - 1][y][j]);
                }
            }
        }
    }
    for(int i = 1; i <= m || i <= n; i ++) {
        idx[i] = 0;
        while((1 << (idx[i] + 1)) <= i) idx[i] ++;
    }
}

int rmq(int lx, int rx, int ly, int ry) {
    int a = idx[rx - lx + 1], la = (1 << a);
    int b = idx[ry - ly + 1], lb = (1 << b);
    return max(max(max(mx[lx][a][ly][b],
                       mx[rx - la + 1][a][ly][b]),
                       mx[lx][a][ry - lb + 1][b]),
                       mx[rx - la + 1][a][ry - lb + 1][b]);
}

int main()
{
    //FIN;
    while(~scanf("%d %d", &m, &n)) {
        mem0(mx);
        rep (i, 1, m) rep (j, 1, n)
            scanf("%d", &mx[i][0][j][0]);
        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
", ma, ma == mx[lx][0][ly][0] || ma == mx[lx][0][ry][0]
                               || ma == mx[rx][0][ly][0] || ma == mx[rx][0][ry][0]
                               ? "yes" : "no");
        }
    }
    return 0;
}