Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 9576 | Accepted: 4506 |
Description
possible to reach through lines every other place, however it need not be a direct connection, it can go through several exchanges. From time to time the power supply fails at a place and then the exchange does not operate. The officials from TLC realized that in such a case it can happen that besides the fact that the place with the failure is unreachable, this can also cause that some other places cannot connect to each other. In such a case we will say the place (where the failure
occured) is critical. Now the officials are trying to write a program for finding the number of all such critical places. Help them.
Input
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;
Output
Sample Input
5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0
Sample Output
1 2
Hint
procedure dfs(u,fa:longint);
var i,count:longint;
begin
inc(time); dfn[u]:=time; low[u]:=time;
vis[u]:=true;
i:=headlist[u];
count:=0;
if ((t[headlist[u]]=fa) and (next[headlist[u]]=-1)) then is_cut[u]:=false;
while i<>-1 do
begin
if not(vis[t[i]]) then begin
dfs(t[i],u);
if low[t[i]]>=dfn[u] then is_cut[u]:=true;
low[u]:=min(low[u],low[t[i]]);
inc(count);
end
else if t[i]<>fa then low[u]:=min(low[u],dfn[t[i]]);
i:=next[i];
end;
if u=1 then is_cut[u]:=false;
if (u=1) and (count>1) then inc(ans);
if is_cut[u] then inc(ans);
end;