傳送門 $dp$，把切割變成往外擴充套件，$f[i][x1][y1][x2][y2]$表示第$i$次擴充套件到$(x1,y1) (x2,y2)$的矩形最小的平方和，然後列舉上一次從哪兒擴充套件過來就好了

#include<iostream>
#include<cstdio>
#include<cstring>
#define LL long long
#define inf 0x3f3f3f3f
using namespace std;
 

int m,a[20][20],sum[20][20];
LL f[20][10][10][10][10];
const int n=8;

inline int rd(){
	int x=0,f=1;char c=' ';
	while(c<'0' || c>'9') f=c=='-'?-1:1,c=getchar();
	while(c<='9' && c>='0') x=x*10+c-'0',c=getchar();
	return x*f;
}

inline void init(){
	m=rd();
	for(int i=1;i<=n;i++)
		for(int
 
 j=1;j<=n;j++)
			a[i][j]=rd();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			sum[i][j]=sum[i][j-1]+a[i][j];
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++)
			sum[i][j]+=sum[i-1][j];
}
inline int min(int x,int y){return x<y?x:y;}
inline LL calc(int x1,int y1,int x2,int y2){
	int
 
 val=sum[x2][y2]-sum[x1-1][y2]-sum[x2][y1-1]+sum[x1-1][y1-1];
	return 1LL*val*val;
}

int main(){
	init(); 
	for(int x1=1;x1<=n;x1++)
		for(int y1=1;y1<=n;y1++)
			for(int x2=x1;x2<=n;x2++)
				for(int y2=y1;y2<=n;y2++)
					f[1][x1][y1][x2][y2]=calc(x1,y1,x2,y2);
	for(int i=2;i<=m;i++)
		for(int x1=1;x1<=n;x1++) 
			for(int y1=1;y1<=n;y1++)
				for(int x2=x1;x2<=n;x2++)
					for(int y2=y1;y2<=n;y2++){
						int tmp=inf;
						for(int j=x1;j<x2;j++)
						tmp=min(tmp,f[i-1][x1][y1][j][y2]+calc(j+1,y1,x2,y2)),
						tmp=min(tmp,f[i-1][j+1][y1][x2][y2]+calc(x1,y1,j,y2));
						for(int j=y1;j<y2;j++)
						tmp=min(tmp,f[i-1][x1][y1][x2][j]+calc(x1,j+1,x2,y2)),
						tmp=min(tmp,f[i-1][x1][j+1][x2][y2]+calc(x1,y1,x2,j));
						f[i][x1][y1][x2][y2]=tmp;
					}
	printf("%lld\n",f[m][1][1][n][n]);
	return 0;
}