Description

将一个８*８的棋盘进行如下分割：将原棋盘割下一块矩形棋盘并使剩下部分也是矩形，再将剩下的部分继续如此分割，这样割了(n-1)次后，连同最后剩下的矩形棋盘共有n块矩形棋盘。(每次切割都只能沿着棋盘格子的边进行) 

原棋盘上每一格有一个分值，一块矩形棋盘的总分为其所含各格分值之和。现在需要把棋盘按上述规则分割成n块矩形棋盘，并使各矩形棋盘总分的均方差最小。 
均方差，其中平均值，x_i为第i块矩形棋盘的总分。 
请编程对给出的棋盘及n，求出O‘的最小值。 

Input

Output

Sample Input

3
1 1 1 1 1 1 1 3
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 0
1 1 1 1 1 1 0 3

Sample Output

1.633

Source

Noi 99

经推倒得出：n个快的平方和最小时候平方差最小

记忆化dp

#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cstdio>
#define N 9
#define M 16
#define INF 0x7ffffff
using namespace std;
int dp[N][N][N][N][M];
bool status[N][N][N][N][M];
int a[N][N],sum[N][N][N][N];
int main()
{
    //freopen("data.txt","r",stdin);
    int dfs(int x1,int y1,int x2,int y2,int n);
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int ss = 0;
        for(int i=1;i<=8;i++)
        {
            for(int j=1;j<=8;j++)
            {
                scanf("%d",&a[i][j]);
                ss+=a[i][j];
            }
        }
        for(int i=1;i<=8;i++)
        {
            for(int j=1;j<=8;j++)
            {
                for(int u=i;u<=8;u++)
                {
                    for(int v=j;v<=8;v++)
                    {
                        int s = 0;
                        for(int w=i;w<=u;w++)
                        {
                            for(int t=j;t<=v;t++)
                            {
                                s+=a[w][t];
                            }
                        }
                        sum[i][j][u][v] = s;
                    }
                }
            }
        }
        memset(status,false,sizeof(status));
        int ans = dfs(1,1,8,8,n);
        double ave = (double)ss/n;
        double res = (double)ans-2*ss*ave+ave*ave*n;
        printf("%.3lf\n",sqrt(res/n));
    }
    return 0;
}
int dfs(int x1,int y1,int x2,int y2,int n)
{
    if(status[x1][y1][x2][y2][n])
    {
        return dp[x1][y1][x2][y2][n];
    }
    if(x1==x2&&y1==y2&&n>1)
    {
        return INF;
    }
    if(n==1)
    {
        return sum[x1][y1][x2][y2]*sum[x1][y1][x2][y2];
    }
    int ans =INF;
    for(int i=x1;i<=x2-1;i++)
    {
        int s1 = dfs(x1,y1,i,y2,n-1);
        if(s1!=INF)
        {
            s1+=sum[i+1][y1][x2][y2]*sum[i+1][y1][x2][y2];
        }
        int s2 = dfs(i+1,y1,x2,y2,n-1);
        if(s2!=INF)
        {
            s2+=sum[x1][y1][i][y2]*sum[x1][y1][i][y2];
        }
        ans = min(ans,s1);
        ans = min(ans,s2);
    }
    for(int i=y1;i<=y2-1;i++)
    {
        int s1 = dfs(x1,y1,x2,i,n-1);
        if(s1!=INF)
        {
            s1+=sum[x1][i+1][x2][y2]*sum[x1][i+1][x2][y2];
        }
        int s2 = dfs(x1,i+1,x2,y2,n-1);
        if(s2!=INF)
        {
            s2+=sum[x1][y1][x2][i]*sum[x1][y1][x2][i];
        }
        ans = min(ans,s1);
        ans = min(ans,s2);
    }
    status[x1][y1][x2][y2][n] = true;
    dp[x1][y1][x2][y2][n] = ans;
    return dp[x1][y1][x2][y2][n];
}