poj 1144

      这道题是一道基础题，要我们找出割点数量（相关概念可以参考http://blog.csdn.net/u011008379/article/details/37996755或《ACM—ICPC程序设计系列 图论及应用》（哈尔滨工业大学出版社）P131）。

      如果一个顶点满足如下条件之一，则是割点：1）v为深搜形成树的根，且至少有两个搜索分支；2）v不为深搜形成树的根，但存在边（u，v），使得dfn[u]<=low[v]。具体判断方式参考下面的代码。

      此外，还要注意一个割点可能属于多个点双连通分量。

      需要注意的地方：1）如果以v作为搜索的根节点时，要注意不要通过与v相关联的边去判断。2）如果用scanf读入数据，注意“%d”之后不要有多余的空格；3）找到割点时，要先记录下来，而不是直接计数，因为可能重复；4）注意只有1个点的特殊情况能否正常处理。

      参考博文：http://www.cnblogs.com/jackge/archive/2013/05/01/3053711.html

      额外的测试数据（来源：http://poj.org/showmessage?message_id=342361）：

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

都输出：1      

      代码（C++）：

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>

#define MAXp 109
#define MAXe 10000
using namespace std;

//#define LOCAL

struct Edge{
    int v;
    int next;   
} edge[MAXe*2];

int head[MAXp],dfn[MAXp],low[MAXp],c,ts,r,cnt;
bool flag[MAXp];

void addEdge(int u,int v)
{
     edge[c].v=v;
     edge[c].next=head[u];
     head[u]=c;
     c++;
}

void dfs(int p,int pre)
{
     int i,v;
     dfn[p]=low[p]=++ts;
     for(i=head[p];i!=-1;i=edge[i].next)
     {
          v=edge[i].v; 
          if(!dfn[v])
          {
              dfs(v,p);
              low[p]=min(low[p],low[v]);
              if(p!=r&&dfn[p]<=low[v])  flag[p]=true;   //一个割点属于多个点双连通分量，需要防止重复计算 
              else if(p==r) cnt++;
          }else if(v!=pre)  low[p]=min(low[p],dfn[v]);
     }
}

int main(int argc, char *argv[])
{
#ifdef LOCAL
   freopen("in.txt","r",stdin);
   freopen("out.txt","w",stdout);
#endif
    int n,u,i,tmp,ans;
    string str;
    while(scanf("%d",&n)&&n!=0)
    {
        memset(head,-1,sizeof(head));
        c=0;                       
        while(scanf("%d",&u)&&u!=0)     
        {
            getline(cin,str);  
            tmp=0;           
            for(i=1;i<str.length();i++) //注意输入时空格的影响，所以从1开始 
            {
                if(str[i]>='0'&&str[i]<='9')  tmp=tmp*10+(str[i]-48);    
                else{                 
                   addEdge(u,tmp);
                   addEdge(tmp,u); 
                   tmp=0;
                }                     
            }
            addEdge(u,tmp);
            addEdge(tmp,u);                      
        }
        ts=cnt=0;
        r=1;
        memset(dfn,0,sizeof(dfn));
        memset(flag,false,sizeof(flag));
        dfs(r,-1);
        if(cnt>1) ans=1;
        else ans=0;         
        for(i=1;i<=n;i++) if(flag[i]==true) ans++;       
        printf("%d\n",ans);                     
    } 
    system("PAUSE");
    return EXIT_SUCCESS;
}
           

题目（ http://poj.org/problem?id=1144）：

Network

Time Limit: 1000MS	Memory Limit: 10000K

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      

Hint

You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.