把一个大矩阵分割成n个矩阵，使它们的方差最小。

g[i,j,k,l,path]表示（i,j） 到 （k,l) 的矩阵分割成path个的最小方差，然后暴力搜索+记忆化

O（n^5) (n<=10) ,无压力水过。

Program b;
const
   maxn=10;
var
   n,m,i,j,k,l,deltax,r:longint;
   delta:double;
   f:array[1..maxn,1..maxn] of longint;
   sum:array[1..maxn,1..maxn,1..maxn,1..maxn] of double;
   g:array[1..maxn,1..maxn,1..maxn,1..maxn,1..maxn] of double;
function min(a,b:double):double;
begin
   if a<b then exit(a) else exit(b);
end;
function dfs(x,y,k,l,path:longint):double;
var
   i,j:longint;
   ans:double;
begin
   if g[x,y,k,l,path]<>0 then exit(g[x,y,k,l,path]);

   if path=1 then
   begin
      g[x,y,k,l,1]:=sum[x,y,k,l]-delta;
      g[x,y,k,l,1]:=g[x,y,k,l,1]*g[x,y,k,l,1];

      exit(g[x,y,k,l,1]);
   end;
   ans:=maxlongint;




   for i:=x to k-1 do
      for j:=1 to path-1 do
         ans:=min(ans,dfs(x,y,i,l,j)+dfs(i+1,y,k,l,path-j));
   for j:=y to l-1 do
      for i:=1 to path-1 do
         ans:=min(ans,dfs(x,y,k,j,i)+dfs(x,j+1,k,l,path-i));


   g[x,y,k,l,path]:=ans;
   exit(ans);

end;
begin
   read(n,m,r);
   deltax:=0;
   for i:=1 to n do
      for j:=1 to m do
      begin
         read(f[i,j]);
         sum[i,j,i,j]:=f[i,j];
         inc(deltax,f[i,j]);

      end;



   delta:=deltax/r;
   for i:=1 to n do
      for j:=1 to m do
      begin
         for k:=i+1 to n do
         begin
            sum[i,j,k,j]:=sum[i,j,k-1,j]+f[k,j];
         end;
         for l:=j+1 to m do
         begin
            sum[i,j,i,l]:=sum[i,j,i,l-1]+f[i,l];
         end;

         for k:=i+1 to n do
            for l:=j+1 to n do
            begin
               sum[i,j,k,l]:=sum[i,j,k-1,l]+sum[i,j,k,l-1]-sum[i,j,k-1,l-1]+f[k,l];
           end;
      end;
   fillchar(g,sizeof(g),0);





   writeln(sqrt(dfs(1,1,n,m,r)/r):2:2);


end.

·