『数学推导·平面DP』「NOI1999」棋盘分割

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。

$=\sqrt \frac{\sum (x_i^2-2*x_i*\overline x+n*\overline x^2)}{n}$
$=\sqrt \frac{\sum x_i^2-2*\sum x_i*\overline x+n*\overline x^2}{n}$
$=\sqrt \frac{\sum x_i^2-2*n*\overline x*\overline x+n*\overline x*\overline x}{n}$
$=\sqrt \frac{\sum x_i^2-n*\overline x*\overline x}{n}$
$=\sqrt{\frac{\sum x_i^2}{n}-\overline x^2}$

推到这里就很简洁明了啦！我们只要求解 $\sum x_i$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;
}