# 【记忆化搜索】bzoj1048 [HAOI2007]分割矩阵

#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;
}
posted @ 2015-03-27 11:12  AutSky_JadeK  阅读(226)  评论(0编辑  收藏  举报