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1191 棋盘分割

2010-05-26 23:50 253 查看

#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;
   // 计算s[x1][y1][x2][y2]
没有平方只是简单的和而已
         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;
                    }

                }
            }
        }
   //
进行初始列的初始化工作不抛弃不放弃继续continue的向前工作
          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 ] / n - average * average;

printf( "%.3llf/n " , sqrt(ans));
}
return   0 ;
}

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标签： iostream include 化工工作

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