[模板]二维ST表

考试yy二维ST表失败导致爆零。

其实和一维的ST表很像...

也是设$f[i][j][p][q]$为以$(i, j)$为左上角，长为$2^p$，宽为$2^q$的矩形的最大值。

算法流程是先把每一行都分别求一遍一维的ST表，然后再把行与行之间合并...

查询和一维ST表类似

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define reg register
inline int read() {
    int res=0;char ch=getchar();bool fu=0;
    while(!isdigit(ch))fu|=(ch=='-'),ch=getchar();
    while(isdigit(ch))res=(res<<3)+(res<<1)+(ch^48),ch=getchar();
    return fu?-res:res;
}

int n, m;
int a[305][305];
int st[505][505][10][10];

inline int query(int x1, int y1, int x2, int y2)
{
    int k1 = log2(x2 - x1 + 1), k2 = log2(y2 - y1 + 1);
    return max(st[x1][y1][k1][k2], max(st[x2-(1<<k1)+1][y1][k1][k2], max(st[x1][y2-(1<<k2)+1][k1][k2], st[x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2])));
}

int main()
{
    freopen("yy.in", "r", stdin);
    freopen("yy.out", "w", stdout);
    n = read(), m = read();
    for (reg int i = 1 ; i <= n ; ++i)
        for (reg int j = 1 ; j <= m ; ++j)
            st[i][j][0][0] = a[i][j] = read();
    for (reg int p = 0 ; p <= 9 ; p ++)
        for (reg int q = 0 ; q <= 9 ; q ++)
            if (p != 0 or q != 0)
                for (reg int i = 1 ; i + (1<<p) - 1 <= n ; i ++)
                    for (reg int j = 1 ; j + (1<<q) - 1 <= m ; j ++)
                        if (!p) st[i][j][p][q] = max(st[i][j][p][q - 1], st[i][j+(1<<(q-1))][p][q - 1]);
                        else st[i][j][p][q] = max(st[i][j][p-1][q], st[i+(1<<(p-1))][j][p-1][q]);
    int q = read();
    while(q--)
    {
        int x1=read(),y1=read(),x2=read(),y2=read();
        printf("%d
", query(x1,y1,x2,y2));
    }
    return 0;
}