矩阵查询最大值，相当于查询的时候一个矩形拆成\(4\)个小矩形，和一维\(ST\)表相差无几
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int INF = 0x3f3f3f3f,N = 255;
inline ll read()
{
ll ret=0;char ch=‘ ‘,c=getchar();
while(!(c>=‘0‘&&c<=‘9‘)) ch=c,c=getchar();
while(c>=‘0‘&&c<=‘9‘) ret=(ret<<1)+(ret<<3)+c-‘0‘,c=getchar();
return ch==‘-‘?-ret:ret;
}
int st[N][N][21][21],n,m,Q,logn[N];
void buildst()
{
int k=0;
for(int i=1;i<=max(n,m);i++)
{
if(1<<k<=i) k++;
logn[i]=k-1;
}
for(int k=0;1<<k<=n;k++)
for(int p=0;1<<p<=m;p++)
for(int i=1;i+(1<<k)-1<=n;i++)
for(int j=1;j+(1<<p)-1<=m;j++)
{
if(!k&&!p) continue;
if(k) st[i][j][k][p]=max(st[i][j][k-1][p],st[i+(1<<(k-1))][j][k-1][p]);
if(p) st[i][j][k][p]=max(st[i][j][k][p-1],st[i][j+(1<<(p-1))][k][p-1]);
}
}
inline int Max(int a,int b,int c,int d){return max(max(a,b),max(c,d));}
inline int query(int xa,int ya,int xb,int yb)
{
int k1=logn[xb-xa+1],k2=logn[yb-ya+1];
return Max(st[xa][ya][k1][k2],st[xa][yb-(1<<k2)+1][k1][k2],
st[xb-(1<<k1)+1][ya][k1][k2],st[xb-(1<<k1)+1][yb-(1<<k2)+1][k1][k2]);
}
int main()
{
n=read(),m=read(); Q=read();
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++) st[i][j][0][0]=read();
buildst();
while(Q--)
{
int xa=read(),ya=read(),xb=read(),yb=read();
printf("%d\n",query(xa,ya,xb,yb));
}
return 0;
}