首先预处理每个F点左右，下一共有多少个F点，然后
对于每个为0的点（R），从这个点开始，一直到这个点
下面第一个R点，这一区间中的min(左),min(右)更新答案。
ps：我估计这道题数据有的格式不对，开始过不去，后来改了读入
就能过了
/**************************************************************
Problem:
User: BLADEVIL
Language: Pascal
Result: Accepted
Time: ms
Memory: kb
****************************************************************/
//By BLADEVIL
var
n, m :longint;
map :array[..,..] of longint;
ans, len1, len, len2 :longint;
left, right, down :array[..,..] of longint;
function max(a,b:longint):longint;
begin
if a>b then max:=a else max:=b;
end;
function min(a,b:longint):longint;
begin
if a>b then min:=b else min:=a;
end;
procedure init;
var
i, j, k :longint;
ss :ansistring;
begin
readln(n,m);
for i:= to n do
begin
readln(ss);
k:=;
for j:= to length(ss) do
if ss[j]<>' ' then
begin
inc(k);
if ss[j]='F' then map[i,k]:= else map[i,k]:=;
end;
end;
for i:= to n do
for j:= to m do
if map[i,j]= then left[i,j]:= else left[i,j]:=left[i,j-]+;
for i:=n downto do
for j:=m downto do
begin
if map[i,j]= then down[i,j]:= else down[i,j]:=down[i+,j]+;
if map[i,j]= then right[i,j]:= else right[i,j]:=right[i,j+]+;
end;
end;
procedure main;
var
i, j, k :longint;
begin
for i:= to n do
for j:= to m do
if map[i,j]= then
begin
len:=;
len1:=maxlongint div ;
len2:=maxlongint div ;
for k:= to down[i+,j] do
begin
len1:=min(len1,left[i+k,j]);
len2:=min(len2,right[i+k,j]);
ans:=max(ans,(len1+len2-)*k);
end;
if len1>=maxlongint div then continue;
inc(len,len1);
ans:=max(ans,(len-)*down[i+,j]);
end;
writeln(ans*);
end;
begin
init;
main;
end.