代码没这么复杂吧...
用DP做最大矩阵和啊...
max:=0;
for i:=1 to n do 
 begin
 for j:=1 to n do b[j]:=0;
 for j:=i to n do 
   begin
   for k:=1 to n do inc(b[j],a[j,k]);
   c:=0;
   for k:=1 to n do 
   begin
   if c>0 then c:=c+b[k] else c:=b[k];
   if c>max then max:=c;
   end;
   end;
 end;