【记忆化搜索】bzoj1048 [HAOI2007]分割矩阵

标准差=√(Σ(xi-xba)²/n)=Σ(xi)²+xba*n-2*xba*sum。只需最小化每个分割出来的矩阵的平方和即可。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#define INF 2000000000
typedef double db;
int mem[11][11][11][11][11],a[11][11],pre[11][11],m,n,K;
int sum;
db xba;
int sqr(int x){return x*x;}
int calc(int x1,int y1,int x2,int y2)
{return pre[x2][y2]-pre[x1-1][y2]-pre[x2][y1-1]+pre[x1-1][y1-1];}
int f(int x1,int y1,int x2,int y2,int K)
{
	if(mem[x1][y1][x2][y2][K]<INF) return mem[x1][y1][x2][y2][K];
	if(K==1) return mem[x1][y1][x2][y2][K]=sqr(calc(x1,y1,x2,y2));
	for(int i=x1;i<x2;++i)
	  for(int j=1;j<K;++j)
	    if((y2-y1+1)*(i-x1+1)>=j&&(y2-y1+1)*(x2-i)>=K-j)
	      mem[x1][y1][x2][y2][K]=min(mem[x1][y1][x2][y2][K],f(x1,y1,i,y2,j)+f(i+1,y1,x2,y2,K-j));
	for(int i=y1;i<y2;++i)
	  for(int j=1;j<K;++j)
	    if((x2-x1+1)*(i-y1+1)>=j&&(x2-x1+1)*(y2-i)>=K-j)
	      mem[x1][y1][x2][y2][K]=min(mem[x1][y1][x2][y2][K],f(x1,y1,x2,i,j)+f(x1,i+1,x2,y2,K-j));
	return mem[x1][y1][x2][y2][K];
}
int main()
{
	scanf("%d%d%d",&n,&m,&K);
	for(int i=1;i<=n;++i)
	  for(int j=1;j<=m;++j)
	    {
	      scanf("%d",&a[i][j]);
	      sum+=a[i][j];
	      pre[i][j]=pre[i][j-1]+a[i][j];
	    }
	xba=(db)sum/(db)K;
	for(int j=1;j<=m;++j)
	  for(int i=1;i<=n;++i)
	    pre[i][j]+=pre[i-1][j];
	memset(mem,0x7f,sizeof(mem));
	printf("%.2f\n",sqrt(((db)f(1,1,n,m,K)+(db)K*xba*xba-2.0*(db)sum*xba)/(db)K));
	return 0;
}