poj 1191 矩形块的划分

思路：黑书的例题

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define Maxn 20
#define mul(a) ((a)*(a))
using namespace std;
int dp[][][][][];
int s[][],val[][];
int S(int x1,int y1,int x2,int y2)
{
    return mul(s[x2][y2]-s[x2][y1-]-s[x1-][y2]+s[x1-][y1-]);
}
int main()
{
    int n,m,i,j;
    double avg;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp,,sizeof(dp));
        memset(s,,sizeof(s));
        for(i=;i<=;i++)
            for(j=;j<=;j++){
            scanf("%d",&val[i][j]);
            s[i][j]=s[i-][j]+s[i][j-]-s[i-][j-]+val[i][j];
        }
        int x1,y1,x2,y2;
        for(x1=;x1<=;x1++)
        for(y1=;y1<=;y1++)
        for(x2=x1;x2<=;x2++)
        for(y2=y1;y2<=;y2++){
            dp[][x1][y1][x2][y2]=S(x1,y1,x2,y2);
        }
        int a,b;
        for(i=;i<=n-;i++)
        for(x1=;x1>=;x1--)
        for(y1=;y1>=;y1--)
        for(x2=x1;x2<=;x2++)
        for(y2=y1;y2<=;y2++){
            int temp;
            for(a=x1;a<x2;a++){
            temp=min(dp[i-][x1][y1][a][y2]+S(a+,y1,x2,y2),dp[i-][a+][y1][x2][y2]+S(x1,y1,a,y2));
            dp[i][x1][y1][x2][y2]=min(temp,dp[i][x1][y1][x2][y2]);
            }
            for(b=y1;b<y2;b++){
            temp=min(dp[i-][x1][y1][x2][b]+S(x1,b+,x2,y2),dp[i-][x1][b+][x2][y2]+S(x1,y1,x2,b));
            dp[i][x1][y1][x2][y2]=min(dp[i][x1][y1][x2][y2],temp);
            }
        }
        double ans;
        avg=s[][]*1.0/(n*1.0);
        ans=(double)dp[n-][][][][];
        ans=ans/(n*1.0);
        ans-=avg*avg;
        printf("%.3lf\n",sqrt(ans));
    }
    return ;
}