poj 1191 棋盘分割 夜

http://poj.org/problem?id=1191

DP 一不小心开了个五维数组

double ans[k][x1][y1][x2][y2]  表示(x1,y1)到(x2,y2)可以割k次时与平均数最小平方和

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<queue>
#include<algorithm>
#include<set>

using namespace std;

const int N=15;
const double M=100000000.0;
double ans[N][10][10][10][10];
double K;
int sum[10][10][10][10];
int a[10][10];
int n;
int dpsum(int x1,int y1,int x2,int y2)//求(x1,y1)到(x2,y2) 的和
{
    if(sum[x1][y1][x2][y2]!=-1)
    return sum[x1][y1][x2][y2];
    if(x1==x2&&y1==y2)
    {
        sum[x1][y1][x2][y2]=a[x1][y1];
    }else
    if(x1<x2)
    {
        sum[x1][y1][x2][y2]=dpsum(x1,y1,x1,y2)+dpsum(x1+1,y1,x2,y2);
    }else
    if(y1<y2)
    {
        sum[x1][y1][x2][y2]=dpsum(x1,y1,x2,y1)+dpsum(x1,y1+1,x2,y2);
    }
    return sum[x1][y1][x2][y2];
}
 double FFmin(double x,double y)
{
    return (x<y)?x:y;
}
double dpans(int k,int x1,int y1,int x2,int y2)
{
    if(ans[k][x1][y1][x2][y2]>=0.0)
    {
        return ans[k][x1][y1][x2][y2];
    }
    if(k==0)
    {
        ans[k][x1][y1][x2][y2]=((sum[x1][y1][x2][y2]-K)*(sum[x1][y1][x2][y2]-K));
        return ans[k][x1][y1][x2][y2];
    }
    ans[k][x1][y1][x2][y2]=M;
    for(int x=x1;x<x2;++x)
    {
       ans[k][x1][y1][x2][y2]=FFmin(ans[k][x1][y1][x2][y2],
                                  (dpans(k-1,x1,y1,x,y2))+(sum[x+1][y1][x2][y2]-K)*(sum[x+1][y1][x2][y2]-K));
       ans[k][x1][y1][x2][y2]=FFmin(ans[k][x1][y1][x2][y2],
                                  (dpans(k-1,x+1,y1,x2,y2))+(sum[x1][y1][x][y2]-K)*(sum[x1][y1][x][y2]-K));
    }
    for(int y=y1;y<y2;++y)
    {
       ans[k][x1][y1][x2][y2]=FFmin(ans[k][x1][y1][x2][y2],
                                  (dpans(k-1,x1,y1,x2,y))+(sum[x1][y+1][x2][y2]-K)*(sum[x1][y+1][x2][y2]-K));
       ans[k][x1][y1][x2][y2]=FFmin(ans[k][x1][y1][x2][y2],
                                  (dpans(k-1,x1,y+1,x2,y2))+(sum[x1][y1][x2][y]-K)*(sum[x1][y1][x2][y]-K));
    }
    return ans[k][x1][y1][x2][y2];
}
void begin(int n)
{
    for(int i=0;i<n;++i)
    for(int l1=1;l1<=8;++l1)
    for(int l2=1;l2<=8;++l2)
    for(int l3=1;l3<=8;++l3)
    for(int l4=1;l4<=8;++l4)
    {
        ans[i][l1][l2][l3][l4]=-1.0;
    }
    for(int l1=1;l1<=8;++l1)
    for(int l2=1;l2<=8;++l2)
    for(int l3=1;l3<=8;++l3)
    for(int l4=1;l4<=8;++l4)
    {
       dpsum(l1,l2,l3,l4);
    }
}
int main()
{
   while(scanf("%d",&n)!=EOF)
   {
       for(int i=1;i<=8;++i)
       {
           for(int j=1;j<=8;++j)
           {
               scanf("%d",&a[i][j]);
           }
       }
       memset(sum,-1,sizeof(sum));
       begin(n);
       K=1.0*sum[1][1][8][8]/n;
       printf("%.3f\n",sqrt(1.0*dpans(n-1,1,1,8,8)/n));
   }
   return 0;
}