poj 1191

这题好像有点卡精度啊。。我最后用c++交，ac了

# include <cstdio>
# include <cstring>
# include <cmath>
# include <iostream>
# define inf 0x3fffffff
using namespace std;
int map[8][8];
double ave,d[20][10][10][10][10],sum[10][10];
double f(int i,int j,int m,int n)
{
    double t = sum[j+1][n+1] - sum[j+1][m] - sum[i][n+1] + sum[i][m];
    return (t - ave)*(t- ave);
}
double dp(int k,int i,int j,int m,int n)
{
    double& res = d[k][i][j][m][n];
    if (k == 1) return f(i,j,m,n);
    if (res>1e-6) return res;
    int c; res = inf;
    for (c=i;c<j;++c)
        res = min(res,min(dp(k-1,i,c,m,n)+f(c+1,j,m,n),dp(k-1,c+1,j,m,n)+f(i,c,m,n)));
    for (c=m;c<n;++c)
        res = min(res,min(dp(k-1,i,j,m,c)+f(i,j,c+1,n),dp(k-1,i,j,c+1,n)+f(i,j,m,c)));
    return res;
}
int main()
{
    int n,i,j,k,m,t,s=0;
    scanf("%d",&n);
    for (i=0;i<8;++i)
        for (j=0;j<8;++j)
        {
            scanf("%d",&map[i][j]);    
            s += map[i][j];
        }
    for (i=1;i<=8;++i)
        for (j=1;j<=8;++j)
            sum[i][j] = sum[i-1][j]+sum[i][j-1]+map[i-1][j-1]-sum[i-1][j-1];    
    ave = s/(n*1.0);
    memset(d,-1,sizeof(d));
     double ans = sqrt(dp(n,0,7,0,7)/n);
    printf("%.3lf
",ans);
    return 0;
}