poj1144 Network 双连通分量

题意/Description:

    A Telephone Line Company (TLC) is establishing a new telephone cable network.
They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange
in each place. From each place it is 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.

读入/Input：


 
      The input file consists of several blocks of lines. Each block describes one network. In the first line of each block there is the number of places N < 100. Each of the next at most N lines contains the number of a place followed by the numbers of some
places to which there is a direct line from this place. These at most N lines completely describe the network, i.e., each direct connection of two places in the network is contained at least in one row. All numbers in one line are separated by
one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;



输出/Output：


 
      The output contains for each block except the last in the input file one line containing the number of critical places.




题解/solution：


 
      改一下输入（输入十分神奇），水水的连通图。详情见：

http://www.cnblogs.com/en-heng/p/4002658.html

代码/Code：


type
arr=record
y,next,op:longint;
f:boolean;
end;
var
dfn,low,ls:array [0..101] of longint;
tu:array [0..1001] of arr;
ans:array [0..101] of boolean;
n,m,tot,p,cnt:longint;
procedure add(o,p:longint);
begin
inc(tot);
with tu[tot] do
begin
y:=p; op:=tot+1;
next:=ls[o];
ls[o]:=tot;
end;
inc(tot);
with tu[tot] do
begin
y:=o; op:=tot-1;
next:=ls[p];
ls[p]:=tot;
end;
end;

function min(o,p:longint):longint;
begin
if o<p then exit(o);
exit(p);
end;

procedure tarjan(x:longint);
var
t:longint;
begin
inc(p);
dfn[x]:=p; low[x]:=p;
t:=ls[x];
while t>0 do
with tu[t] do
begin
if not f then
begin
f:=true;
tu[op].f:=true;
if dfn[y]=0 then
begin
if x=1 then inc(cnt);
tarjan(y);
low[x]:=min(low[x],low[y]);
if low[y]>=dfn[x] then ans[x]:=true;
end else
low[x]:=min(low[x],dfn[y]);
end;
t:=next;
end;
end;

procedure print;
var
i,sum:longint;
begin
if cnt>=2 then ans[1]:=true
else ans[1]:=false;
sum:=0;
for i:=1 to n do
if ans[i] then inc(sum);
writeln(sum);
end;

procedure init;
var
i,x,y:longint;
begin
readln(n);
while n<>0 do
begin
cnt:=0; tot:=0;
fillchar(dfn,sizeof(dfn),0);
fillchar(low,sizeof(low),0);
fillchar(tu,sizeof(tu),0);
fillchar(ls,sizeof(ls),0);
fillchar(ans,sizeof(ans),0);
read(x);
while x<>0 do
begin
while not eoln do
begin
read(y);
add(x,y);
end;
read(x);
end;
tarjan(1);
print;
readln(n);
end;
end;

begin
init;
end.