做法：
设立数组f[x1][y1][x2][y2][k]表示在矩形x1y1,x2y2中还需切割k-1次；
逐步递推即可。
#include<stdio.h>
#include<math.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<stdlib.h>
#define INF_MAX 0x7fffffff
#define INF 999999
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
struct node
{
int u;
int v;
int w;
bool friend operator < (node a, node b){
return a.w < b.w;
}
}edge[1001];
int gcd(int n,int m){if(n<m) swap(n,m);return n%m==0?m:gcd(m,n%m);}
int lcm(int n,int m){if(n<m) swap(n,m);return n/gcd(n,m)*m;}
int f[10][10][10][10][21];
int p[10][10][10][10];
int a[10][10];
int init(int x1,int y1,int x2,int y2)
{
if(p[x1][y1][x2][y2]>=0)return p[x1][y1][x2][y2];
int sum;
sum=0;
int i,j;
for(i=x1;i<=x2;i++)
{
for(j=y1;j<=y2;j++)
{
sum+=a[i][j];
}
}
p[x1][y1][x2][y2]=sum*sum;
// printf("-----%d %d %d %d %d\n",x1,y1,x2,y2,sum*sum);
return sum*sum;
}
int dos(int x1,int y1,int x2,int y2,int k)
{
int i;
if(k==1)return init(x1,y1,x2,y2);
if(f[x1][y1][x2][y2][k]>=0)return f[x1][y1][x2][y2][k];
int mins=99999999;
for(i=x1;i<x2;i++)
{
mins=min3(mins,(dos(i+1,y1,x2,y2,k-1)+init(x1,y1,i,y2)),(init(i+1,y1,x2,y2)+dos(x1,y1,i,y2,k-1)));
}
for(i=y1;i<y2;i++)
{
mins=min3(mins,(dos(x1,i+1,x2,y2,k-1)+init(x1,y1,x2,i)),(init(x1,i+1,x2,y2)+dos(x1,y1,x2,i,k-1)));
}
f[x1][y1][x2][y2][k]=mins;
//printf("%d %d %d %d %d %d\n",x1,y1,x2,y2,k,mins);
return mins;
}
int main()
{
int n,i,j,ans;
scanf("%d",&n);
double x;
x=0;
mem(f,-1);
mem(p,-1);
for(i=1;i<=8;i++)
{
for(j=1;j<=8;j++)
scanf("%d",&a[i][j]);
}
x=1.0*x/n;
x=x*x;
ans=dos(1,1,8,8,n);
x=(sqrt(1.0*init(1,1,8,8))/n);
printf("%.3f\n",sqrt(1.0*ans/n-x*x));
return 0;
}