考虑对于任意两个数q1[i]和q1[j]来说,它们不能压入同一个栈中的充要条件是存在一个k,使得i<j<k且q1[k]<q1[i]<q1[j] 我们对所有的数对(i,j)满足1<=i<j<=n,检查是否存在i<j<k满足p1[k]<p1[i]<p1[j].如果存在,那么在点i和点j之间连一条无向边,表示p1[i]和p1[j]不能压入同一个栈，这就是二分图。只要判断这是不是二分图，如果是，模拟输出即可
program twostack;
var
n,i:integer;
s:array[1..2,0..1000] of integer;
a,color:array[1..1000] of integer;
b:array[1..1001] of integer;
g:array[1..1000,0..1000] of integer;
procedure build;
var
i,j,min:integer;
begin
min:=maxint;
for i:=n downto 1 do
begin
if a[i]<min then min:=a[i];
b[i]:=min;
end;
b[n+1]:=n+1;
for i:=1 to n-1 do
for j:=i+1 to n do
if (b[j+1]<a[i]) and (a[i]<a[j]) then
begin
inc(g[i,0]);
g[i,g[i,0]]:=j;
inc(g[j,0]);
g[j,g[j,0]]:=i;
end;
end;
procedure dfs(i:integer);
var
j:integer;
begin
for j:=1 to g[i,0] do
if color[g[i,j]]=0 then
begin
color[g[i,j]]:=3-color[i];
dfs(g[i,j]);
end
else if color[g[i,j]]=color[i] then
begin
writeln('0');
close(input);
close(output);
halt;
end;
end;
procedure print;
var
i,now,c:integer;
begin
now:=1;
for i:=1 to n do
begin
if color[i]=1 then write('a ')
else write('c ');
c:=color[i];
inc(s[c,0]);
s[c,s[c,0]]:=a[i];
while (s[1,s[1,0]]=now) or (s[2,s[2,0]]=now) do
begin
if s[1,s[1,0]]=now then
begin
if now<>n then write('b ')
else writeln('b');
dec(s[1,0]);
end
else
begin
if now<>n then write('d ')
else writeln('d');
dec(s[2,0]);
end;
inc(now);
end;
end;
end;
begin
assign(input,'twostack.in');
reset(input);
assign(output,'twostack.out');
rewrite(output);
fillchar(a,sizeof(a),0);
fillchar(b,sizeof(b),0);
fillchar(g,sizeof(g),0);
fillchar(s,sizeof(s),0);
fillchar(color,sizeof(color),0);
readln(n);
for i:=1 to n do
read(a[i]);
readln;
build;
for i:=1 to n do
if color[i]=0 then
begin
color[i]:=1;
dfs(i);
end;
print;
close(input);
close(output);
end.
