POJ 1144 无向图求割点

                                                                                  Network

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7966		Accepted: 3755

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.

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;

Output

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

Sample Input

5
5 1 2 3 4
0
6
2 1 3
5 4 6 2
0
0

Sample Output

1
2

题意：求割点的个数；

割点模板：

#include<iostream>
#include<cstring>
#include<stdio.h>
#include<algorithm>
#include<string>
using namespace std;
struct node
{
         int to,next;
}edge[10000];
int head[110],tol,n,low[110],dfn[110],iscut[110],indexx,cut;
void add(int u,int v)
{
         edge[tol].to=v;
         edge[tol].next=head[u];
         head[u]=tol++;
}
void tarjin(int u,int fa)
{
         low[u]=dfn[u]=++indexx;
         int i,v,son=0;
         for(i=head[u];i!=-1;i=edge[i].next)
         {
                  v=edge[i].to;
                  if(!dfn[v])
                  {
                           son++;
                           tarjin(v,u);
                           if(low[u]>low[v])low[u]=low[v];
                           if(low[v]>=dfn[u])iscut[u]=1;
                  }
                  else if(v!=fa&&low[u]>dfn[v])low[u]=dfn[v];
         }
         if(fa<0&&son==1)iscut[u]=0;
}
void solve()
{
         memset(dfn,0,sizeof(dfn));
         memset(iscut,0,sizeof(iscut));
         indexx=0;
         cut=0;
         tarjin(1,-1);
         for(int i=1;i<=n;i++)cut+=iscut[i];
         cout<<cut<<endl;
}
int main()
{
         int i,j,k,m;char ch;
         while(~scanf("%d",&n)&&n)
         {
                  memset(head,-1,sizeof(head));
                  tol=0;
                  while(~scanf("%d",&i)&&i)
                  {
                           while((ch=getchar())!='\n')
                           {
                                    scanf("%d",&j);
                                    add(i,j);
                                    add(j,i);
                           }
                  }
                 // cout<<tol<<endl;
                  solve();
         }
         return 0;
}