\(Solution:\)
直接大力预处理,\(O(n^2log_2n)+O(q)\)可过
\(Code:\)
#include<bits/stdc++.h>
using namespace std;
namespace my_std
{
typedef long long ll;
typedef double db;
const db PI=acos(-1.0);
#define cp complex<db>
#define MP make_pair
#define fir first
#define sec second
#define fr(i,x,y) for(int i=(x);i<=(y);i++)
#define pfr(i,x,y) for(int i=(y);i>=(x);i--)
#define gfr(u) for(int i=head[u];i;i=e[i].nxt)
#define pf printf
inline ll read()
{
ll sum=0,f=1;
char ch=0;
while(!isdigit(ch))
{
if(ch=='-') f=-1;
ch=getchar();
}
while(isdigit(ch))
{
sum=(sum<<1)+(sum<<3)+(ch^48);
ch=getchar();
}
return sum*f;
}
inline void write(int x)
{
if(x<0)
{
putchar('-');
x=-x;
}
if(x>9) write(x/10);
putchar(x%10+'0');
}
}
using namespace my_std;
const int N=1050;
int n,m,q,a[N][N],b[N][N],aa[N][N],bb[N][N],ans[N][N];
int main(void)
{
n=read(),m=read(),q=read();
fr(i,1,n) fr(j,1,m) a[i][j]=read();
fr(i,1,m) fr(j,1,n) b[i][j]=a[j][i];
fr(i,1,n) fr(j,1,m) aa[i][j]=a[i][j];
fr(i,1,m) fr(j,1,n) bb[i][j]=b[i][j];
fr(i,1,n) sort(aa[i]+1,aa[i]+m+1);
fr(i,1,m) sort(bb[i]+1,bb[i]+n+1);
fr(i,1,n)
{
fr(j,1,m)
{
int x=lower_bound(aa[i]+1,aa[i]+m+1,a[i][j])-aa[i];
int y=lower_bound(bb[j]+1,bb[j]+n+1,a[i][j])-bb[j];
ans[m-x+1][n-y+1] ++;
}
}
while(q--)
{
int x=read(),y=read();
write(ans[x][y]);
putchar('\n');
}
return 0;
}