poj 1191 棋盘分割
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;
}