只是注意本题数据定义小一点,不然会超内存。
#include <stdio.h>
#include <iostream>
#include <math.h>
using namespace std;
const int N = 255;
int n, b, k;
short val[N][N];
short dpmax[N][N][10][10];
short dpmin[N][N][10][10];
void ST()
{
int i, j, r, c, k;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
dpmax[i][j][0][0]=dpmin[i][j][0][0]=val[i][j];
k=(int)(log(double(n))/log(2.0));
for(i=0;i<=k;i++)
{
for(j=0;j<=k;j++)
{
if(i==0&&j==0) continue;
for(r=1;r+(1<<i)-1<=n;r++)
{
for(c=1;c+(1<<j)-1<=n;c++)
{
if(i==0)
{
dpmax[r][c][i][j]=max(dpmax[r][c][i][j-1],dpmax[r][c+(1<<(j-1))][i][j-1]);
dpmin[r][c][i][j]=min(dpmin[r][c][i][j-1],dpmin[r][c+(1<<(j-1))][i][j-1]);
}
else
{
dpmax[r][c][i][j]=max(dpmax[r][c][i-1][j],dpmax[r+(1<<(i-1))][c][i-1][j]);
dpmin[r][c][i][j]=min(dpmin[r][c][i-1][j],dpmin[r+(1<<(i-1))][c][i-1][j]);
}
}
}
}
}
}
short query(int r1, int c1, int r2, int c2)
{
int kr=(int)(log(double(r2-r1+1))/log(2.0));
int kc=(int)(log(double(c2-c1+1))/log(2.0));
short t1=dpmax[r1][c1][kr][kc];
short t2=dpmax[r2-(1<<kr)+1][c1][kr][kc];
short t3=dpmax[r1][c2-(1<<kc)+1][kr][kc];
short t4=dpmax[r2-(1<<kr)+1][c2-(1<<kc)+1][kr][kc];
short m1=dpmin[r1][c1][kr][kc];
short m2=dpmin[r2-(1<<kr)+1][c1][kr][kc];
short m3=dpmin[r1][c2-(1<<kc)+1][kr][kc];
short m4=dpmin[r2-(1<<kr)+1][c2-(1<<kc)+1][kr][kc];
return max(max(t1,t2),max(t3,t4))-min(min(m1,m2),min(m3,m4));
}
int main()
{
int i, j;
int r1,c1,r2,c2;
while(~scanf("%d%d%d",&n,&b,&k))
{
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&val[i][j]);
ST();
while(k--)
{
scanf("%d%d",&r1,&c1);
r2=r1+b-1;
c2=c1+b-1;
printf("%d\n",query(r1,c1,r2,c2));
}
}
return 0;
}