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/*
dp[k][x1][y1][x2][y2]:左上角坐标为(x1,y1),右下角坐标为(x2,y2)
的棋盘,设它把切割k次以后得到的k+1块矩形的总分平方和最小值.

s[x1][y1][x2][y2]:左上角坐标为(x1,y1),右下角坐标为(x2,y2)
的棋盘的总和的平方


 dp[k][x1][y1][x2][y2] =
1)按横的划分: min(dp[k-1][x1][y1][f][y2]+s[f+1][y1][x2][y2]
    , dp[k-1][f+1][y1][x2][y2]+s[x1][y1][f][y2]);

2)按竖的划分: min(dp[k-1][x1][y1][x2][f]+s[x1][f+1][x2][y2]
    , dp[k-1][x1][f+1][x2][y2]+s[x1][y1][x2][f]);
*/

#include  < iostream >
#include 
< cmath >
#define  INF 10000000
using   namespace  std;

const   int  m  =   8 ;
int  a[ 10 ][ 10 ];
int  n;
double  dp[ 16 ][ 10 ][ 10 ][ 10 ][ 10 ], s[ 10 ][ 10 ][ 10 ][ 10 ];

/*
dp[k][x1][y1][x2][y2]:左上角坐标为(x1,y1),右下角坐标为(x2,y2)
的棋盘,设它把切割k次以后得到的k+1块矩形的总分平方和最小值.

s[x1][y1][x2][y2]:左上角坐标为(x1,y1),右下角坐标为(x2,y2)
的棋盘的总和的平方


 dp[k][x1][y1][x2][y2] = 
1)按横的划分:    min(dp[k-1][x1][y1][f][y2]+s[f+1][y1][x2][y2]
                , dp[k-1][f+1][y1][x2][y2]+s[x1][y1][f][y2]);

2)按竖的划分:    min(dp[k-1][x1][y1][x2][f]+s[x1][f+1][x2][y2]
                , dp[k-1][x1][f+1][x2][y2]+s[x1][y1][x2][f]);
*/


double  mmin( double  a,  double  b)
{
    
return  a  <  b  ?  a: b;
}


int  main()
{
    
// freopen("data.txt", "r", std);

    
int  x1, y1, x2, y2, t, f;
    
double  sum, average, temp;
    
while (cin  >>  n) {
        sum 
=   0 ;
        
for (x1  =   0 ; x1  <  m; x1 ++ )
            
for (y1  =   0 ; y1  <  m; y1 ++ ) {
                cin 
>>  a[x1][y1];
                sum 
+=  a[x1][y1];
            }

        average 
=  sum / n;

        
for (x1  =   0 ; x1  <  m; x1 ++ ) {
            
for (y1  =   0 ; y1  <  m; y1 ++ ) {
                
for (x2  =  x1; x2  <  m; x2 ++ ) {
                    sum 
=   0 ;
                    
for (y2  =  y1; y2  <  m; y2 ++ ) {
                        sum 
+=  a[x2][y2];
                        
if (x1  ==  x2)
                            s[x1][y1][x2][y2] 
=  sum;
                        
else
                            s[x1][y1][x2][y2] 
=  s[x1][y1][x2 - 1 ][y2]  +  sum;
                    }

                }

            }

        }



        
for (x1  =   0 ; x1  <  m; x1 ++ ) {
            
for (y1  =   0 ; y1  <  m; y1 ++ ) {
                
for (x2  =  x1; x2  <  m; x2 ++ ) {
                    
for (y2  =  y1; y2  <  m; y2 ++ ) {
                        s[x1][y1][x2][y2] 
*=  s[x1][y1][x2][y2];
                        dp[
0 ][x1][y1][x2][y2]  =  s[x1][y1][x2][y2];
                    }

                }

            }

        }


        
for (t  =   1 ; t  <  n; t ++ ) {
            
for (x1  =   0 ; x1  <  m; x1 ++ ) {
                
for (y1  =   0 ; y1  <  m; y1 ++ ) {
                    
for (x2  =  x1; x2  <  m; x2 ++ ) {
                        
for (y2  =  y1; y2  <  m; y2 ++ ) {
                            dp[t][x1][y1][x2][y2] 
=  INF;
                            
for (f  =  x1; f  <  x2; f ++ ) {
                                temp 
=  mmin(dp[t - 1 ][x1][y1][f][y2] + s[f + 1 ][y1][x2][y2]
                                    , dp[t
- 1 ][f + 1 ][y1][x2][y2] + s[x1][y1][f][y2]);
                                dp[t][x1][y1][x2][y2] 
=  mmin(temp, dp[t][x1][y1][x2][y2]);
                            }

                            
for (f  =  y1; f  <  y2; f ++ ) {
                                temp 
=  mmin(dp[t - 1 ][x1][y1][x2][f] + s[x1][f + 1 ][x2][y2]
                                    , dp[t
- 1 ][x1][f + 1 ][x2][y2] + s[x1][y1][x2][f]);
                                dp[t][x1][y1][x2][y2] 
=  mmin(temp, dp[t][x1][y1][x2][y2]);
                            }

                        }

                    }

                }

            }

        }

        
        
double  ans;
        ans 
=  dp[n - 1 ][ 0 ][ 0 ][m - 1 ][m - 1 ] / -  average * average;
        printf(
" %.3llf\n " , sqrt(ans));
    }

return   0 ;
}
posted on 2009-04-21 19:57 longshen 阅读(1578) 评论(0)  编辑 收藏 引用 所属分类: poj

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