【codevs1049】

没被使用的ID谁都有可能会用到；

而当匹配后，自己的ID还可以和其他人的换时，

即并不能确定两个ID和两个名字的关系时，这两个人的ID均未知。

var
w:array[1..20,1..20] of boolean;
ID:array[1..20] of string;
nm,ans:array[1..20] of string;
room2,used:array[1..20] of boolean;
r,tr:array[1..20] of longint;
N,i,j,p,tmp:longint;
cc:char;
ss:string;
function checkname(s:string):longint;
var
i:longint;
begin
for i:=1 to p do
  if nm[i]=s then
   exit(i);
inc(p);
nm[p]:=s;
checkname:=p;
end;
function checkID(s:string):longint;
var
i:longint;
begin
for i:=1 to n do
  if ID[i]=s then
   exit(i);
end;
function dfs(p:longint):boolean;
var
i:longint;
begin
if p=0 then
  exit(false);
for i:=1 to n do
  if w[p,i] and not(used[i]) then
  begin
   used[i]:=true;
   if (r[i]=0) or dfs(r[i]) then
   begin
    r[i]:=p;
    exit(true);
   end;
  end;
dfs:=false;
end;
begin
readln(N);
for i:=1 to n do
  for j:=1 to n do
   w[i,j]:=true;
for i:=1 to n-1 do
begin
  setlength(ID[i],20);
  read(cc);j:=0;
  while cc<>' ' do
  begin
   inc(j);
   ID[i][j]:=cc;
   read(cc);
  end;
  setlength(ID[i],j);
end;
readln(ID[n]);
readln(ss);
while ss<>'Q' do
begin
  case ss[1] of
  'E':begin
    delete(ss,1,2);
    room2[checkname(ss)]:=true;
   end;
  'L':begin
    delete(ss,1,2);
    room2[checkname(ss)]:=false;
   end;
  'M':begin
    delete(ss,1,2);
    tmp:=checkID(ss);
    for i:=1 to n do
     if not room2[i] then
     begin
      w[tmp,i]:=false;
     //writeln('del ',tmp,' ',i); //debug
     end;
   end;
  end;
  readln(ss);
end;
for i:=1 to n do
begin
  fillchar(used,sizeof(used),0);
  dfs(i);
end;
for i:=1 to n do
  if r[i]>0 then
  begin
   tr:=r;
   w[r[i],i]:=false;
   r[i]:=0;
   fillchar(used,sizeof(used),0);
   if dfs(tr[i]) then
    ans[i]:=nm[i]+':???'
   else
    ans[i]:=nm[i]+':'+ID[tr[i]];
   r:=tr;
   w[r[i],i]:=true;
  end;
for i:=1 to n-1 do
  for j:=i+1 to n do
   if ans[i]>ans[j] then
   begin
    ss:=ans[i];
    ans[i]:=ans[j];
    ans[j]:=ss;
   end;
for i:=1 to n do
  writeln(ans[i]);
end.

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