思路：
先来想想暴力的写法：
(n^{2})枚举左上角的顶点，(k^{2})求最小值。
考虑优化：
(1.)答案有单调性，可以二分答案，省去枚举左上角顶点的复杂度。
(2.)每次(check)的时候，将大于该数的设为(1)，小于该数的设为(0)，这样就可以在(k^{2})枚举的时候用二维前缀和(O(1))查询了。
注意下边界问题，用第几小或第几大都是能过的。
代码：
第几小
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
ll res = 1;
while(b)
{
if(b & 1)res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=3e5+100;
const double eps=1e-7;
ll n,k,a[810][810],s[810][810],idx;
bool check(ll x){
memset(s,0,sizeof s);
rep(i,1,n)
rep(j,1,n)
if(a[i][j]<=x) s[i][j]=1;
else s[i][j]=0;
rep(i,1,n)
rep(j,1,n)
s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+s[i][j];
for(int i=1;i+k-1<=n;i++){
for(int j=1;j+k-1<=n;j++){
ll x2=i+k-1,y2=j+k-1;
ll x1=i,y1=j;
ll sum=s[x2][y2]+s[x1-1][y1-1]-s[x1-1][y2]-s[x2][y1-1];
if(sum>=idx) return 1;
}
}
return 0;
}
int main(){
n=read,k=read;
idx=(k*k);
if(idx%2) idx=idx/2+1;
else idx=idx/2;
ll l=1e9+1,r=-1,res;
rep(i,1,n)
rep(j,1,n){
a[i][j]=read;
l=min(l,a[i][j]);
r=max(r,a[i][j]);
}
while(l<=r){
ll mid=(l+r)/2;
if(check(mid)) r=mid-1,res=mid;
else l=mid+1;
//cout<<l<<"*****"<<r<<"*****"<<mid<<"
";
}
printf("%lld
",res);
return 0;
}
第几大
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
ll x = 0, f = 1;
char ch = getchar();
while(ch < '0' || ch > '9')
{
if(ch == '-')f = -1;
ch = getchar();
}
while(ch >= '0' && ch <= '9')
{
x = x * 10 + ch - '0';
ch = getchar();
}
return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
ll res = 1;
while(b)
{
if(b & 1)res = res * a % p;
a = a * a % p;
b >>= 1;
}
return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)
const int maxn=3e5+100;
const double eps=1e-7;
ll n,k,a[810][810],s[810][810],idx;
bool check(ll x){
memset(s,0,sizeof s);
rep(i,1,n)
rep(j,1,n)
if(a[i][j]>x) s[i][j]=1;
else s[i][j]=0;
rep(i,1,n)
rep(j,1,n)
s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+s[i][j];
for(int i=1;i+k-1<=n;i++){
for(int j=1;j+k-1<=n;j++){
ll x2=i+k-1,y2=j+k-1;
ll x1=i,y1=j;
ll sum=s[x2][y2]+s[x1-1][y1-1]-s[x1-1][y2]-s[x2][y1-1];
if(sum<idx) return 1;
}
}
return 0;
}
int main(){
n=read,k=read;
idx=(k*k)/2+1;
ll l=1e9+1,r=-1,res;
rep(i,1,n)
rep(j,1,n){
a[i][j]=read;
l=min(l,a[i][j]);
r=max(r,a[i][j]);
}
while(l<=r){
ll mid=(l+r)/2;
if(check(mid)) r=mid-1,res=mid;
else l=mid+1;
//cout<<l<<"*****"<<r<<"*****"<<mid<<"
";
}
printf("%lld
",res);
return 0;
}