挑战（NDK1332）

program NDK1332;

const
 maxn=250000;
 
type
 atp=record
  x,y,time:longint;
 end;

var
 n,ans,st,ed:longint;
 fx:char;
 map:array [0..maxn] of atp;
 
procedure init;
var
 i,x,y:longint;
 ch:char;
begin
 readln(n);
 x:=0;
 y:=0;
 for i:=1 to n do 
  begin
   read(ch);
   case ch of
    'N':dec(x);
    'E':inc(y);
    'W':dec(y);
    'S':inc(x);
   end;{case}
   map[i].x:=x;
   map[i].y:=y;
   map[i].time:=i;
  end;
end;

{qsort是先按照一个关键字排序，再按照另一个关键字排序。这样能表示一个关键字相等的情况下比较另一个关键字的大小}
procedure qsort1(l,r:longint);
var
 i,j:longint;
 m,t:atp;
begin
 i:=l;
 j:=r;
 m:=map[(l+r) shr 1];
 repeat
  while (map[i].x<m.x) or ((map[i].x=m.x) and (map[i].y<m.y)) do inc(i);
  while (map[j].x>m.x) or ((map[j].x=m.x) and (map[j].y>m.y)) do dec(j);
  if i<=j then
   begin
    t:=map[i];
    map[i]:=map[j];
    map[j]:=t;
    inc(i);   
    dec(j);
   end;
 until i>j;
 if i<r then qsort1(i,r);
 if l<j then qsort1(l,j);
end;

procedure qsort2(l,r:longint);
var
 i,j:longint;
 m,t:atp;
begin
 i:=l;
 j:=r;
 m:=map[(l+r) shr 1];
 repeat
  while (map[i].y<m.y) or ((map[i].y=m.y) and (map[i].x<m.x)) do inc(i);
  while (map[j].y>m.y) or ((map[j].y=m.y) and (map[j].x>m.x)) do dec(j);
  if i<=j then
   begin
    t:=map[i];
    map[i]:=map[j];
    map[j]:=t;
    inc(i);   
    dec(j);
   end;
 until i>j;
 if i<r then qsort2(i,r);
 if l<j then qsort2(l,j);
end;

procedure main;
var
 i:longint;
begin
 st:=maxlongint;
 ed:=-maxlongint;
 ans:=maxlongint;
 
 qsort1(1,n);
 for i:=1 to n do 
  begin
   if (map[i].x=map[i-1].x) and (abs(map[i].time-map[i-1].time)<>1) then{处于一行或一列同时访问序不能相差1。}
    begin
     if abs(map[i].y-map[i-1].y)<=ans then
      begin
       if abs(map[i].y-map[i-1].y)=ans then{能进行更新的条件。}
        begin
         if map[i].time<=map[i-1].time then
          begin
           if (map[i].time<st) or ((map[i].time=st) and (map[i-1].time>ed)) then
            begin
             st:=map[i].time;
             ed:=map[i-1].time;
             fx:='W';
            end;
          end
         else
          begin
           if (map[i-1].time<st) or ((map[i-1].time=st) and (map[i].time>ed)) then
            begin
             st:=map[i-1].time;
             ed:=map[i].time;
             fx:='E';
            end;
          end;
        end
       else 
        begin
         if map[i].time<map[i-1].time then 
          begin
           st:=map[i].time;
           ed:=map[i-1].time;
           fx:='W';          
          end
         else
          begin
           st:=map[i-1].time;
           ed:=map[i].time;
           fx:='E';
          end;
        end;
       ans:=abs(map[i].y-map[i-1].y); 
      end;     
    end;  
  end;
  
 qsort2(1,n);
 for i:=1 to n do 
  begin
   if (map[i].y=map[i-1].y) and (abs(map[i].time-map[i-1].time)<>1) then
    begin
     if abs(map[i].x-map[i-1].x)<=ans then
      begin
       if abs(map[i].x-map[i-1].x)=ans then
        begin
         if map[i].time<=map[i-1].time then
          begin
           if (map[i].time<st) or ((map[i].time=st) and (map[i-1].time>ed)) then
            begin
             st:=map[i].time;
             fx:='N';
            end;
          end
         else
          begin
           if (map[i-1].time<st) or ((map[i-1].time=st) and (map[i].time>ed)) then
            begin
             st:=map[i-1].time;
             ed:=map[i].time;
             fx:='S';
            end;
          end;
        end
       else 
        begin
         if map[i].time<map[i-1].time then 
          begin
           st:=map[i].time;
           ed:=map[i-1].time;
           fx:='W';          
          end
         else
          begin
           st:=map[i-1].time;
           ed:=map[i].time;
           fx:='E';
          end;
        end;
       ans:=abs(map[i].x-map[i-1].x); 
      end;     
    end;  
  end;  
  
end;


begin
 assign(input,'NDK1332.in'); reset(input);
 assign(output,'NDK1332.out'); rewrite(output); 

 init;
 main;
 writeln(ans,' ',st,' ',ed,' ',fx);
 
 close(input); close(output);
end.