【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
);
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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