bzoj1048 分割矩阵 记忆化搜索

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double  f[15][15][15][15][15];
int a[15][15],s[15][15];
double ave;
double dfs(int a,int b,int c,int d,int k){
double tmp;
if (f[a][b][c][d][k]!=-1) return f[a][b][c][d][k];
if (k==1){
tmp=s[c][d]+s[a-1][b-1]-s[c][b-1]-s[a-1][d];
tmp=(tmp-ave)*(tmp-ave);
return f[a][b][c][d][k]=tmp;
}
int i,j; tmp=1e9;
for (i=a; i<c; i++)
for (j=1; j<k; j++)
tmp=min(tmp,dfs(a,b,i,d,j)+dfs(i+1,b,c,d,k-j));
for (i=b; i<d; i++)
for (j=1; j<k; j++)
tmp=min(tmp,dfs(a,b,c,i,j)+dfs(a,i+1,c,d,k-j));
return f[a][b][c][d][k]=tmp;
}
int main(){
int n,m,k;
scanf("%d%d%d",&n,&m,&k);
int i,j,x,y,t;
for (i=1; i<=n; i++)
for (j=1; j<=m; j++){
scanf("%d",&x);
s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+x;
}
for (i=0; i<=n+1; i++)
for (j=0; j<=m+1; j++)
for (x=0; x<=n+1; x++)
for (y=0; y<=m+1; y++)
for (t=0; t<=k+1; t++)
f[i][j][x][y][t]=-1;
ave=(double)s
[m]/k;
printf("%.2f",sqrt(dfs(1,1,n,m,k)/k));
return 0;
}