题目描述
题解
这道题难就难在这一个均方差公式怎么搞,我们不妨对这一个算式进行化简。
推到这里就很简洁明了啦!我们只要求解 的最小值即可。也就是每一个区域的平方和最小,这个用平面DP的方法可以很简单的得到,这里就不说了。
代码如下:
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int N = 10;
int n = 8, m;
int a[N][N], f[N][N][N][N][20];
int s(int A,int b,int c,int d)
{
int sum = a[c][d] - a[A-1][d] - a[c][b-1] + a[A-1][b-1];
return sum * sum;
}
int main(void)
{
freopen("input.in","r",stdin);
freopen("output.out","w",stdout);
cin >> m;
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
cin>>a[i][j], a[i][j] = a[i-1][j]+a[i][j-1]-a[i-1][j-1]+a[i][j];
for (int r=1;r<=m;++r)
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
for (int k=i;k<=n;++k)
for (int t=j;t<=n;++t)
{
if (r == 1) { f[i][j][k][t][r] = s(i,j,k,t); continue; };
f[i][j][k][t][r] = 1e9;
for (int p=i;p<k;++p) f[i][j][k][t][r] = min(f[i][j][k][t][r],f[i][j][p][t][r-1]+s(p+1,j,k,t));
for (int p=i;p<k;++p) f[i][j][k][t][r] = min(f[i][j][k][t][r],f[p+1][j][k][t][r-1]+s(i,j,p,t));
for (int p=j;p<t;++p) f[i][j][k][t][r] = min(f[i][j][k][t][r],f[i][j][k][p][r-1]+s(i,p+1,k,t));
for (int p=j;p<t;++p) f[i][j][k][t][r] = min(f[i][j][k][t][r],f[i][p+1][k][t][r-1]+s(i,j,k,p));
}
int ans = f[1][1][8][8][m];
int sum = s(1,1,8,8);
printf("%.3lf",sqrt(ans*1.0/(m*1.0)-sum*1.0/(m*m*1.0)));
return 0;
}