poj 1144

      這道題是一道基礎題，要我們找出割點數量（相關概念可以參考http://blog.csdn.net/u011008379/article/details/37996755或《ACM—ICPC程式設計系列 圖論及應用》（哈爾濱工業大學出版社）P131）。

      如果一個頂點滿足如下條件之一，則是割點：1）v為深搜形成樹的根，且至少有兩個搜尋分支；2）v不為深搜形成樹的根，但存在邊（u，v），使得dfn[u]<=low[v]。具體判斷方式參考下面的代碼。

      此外，還要注意一個割點可能屬于多個點雙連通分量。

      需要注意的地方：1）如果以v作為搜尋的根節點時，要注意不要通過與v相關聯的邊去判斷。2）如果用scanf讀入資料，注意“%d”之後不要有多餘的空格；3）找到割點時，要先記錄下來，而不是直接計數，因為可能重複；4）注意隻有1個點的特殊情況能否正常處理。

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

      額外的測試資料（來源：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

都輸出：1      

      代碼（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;   //一個割點屬于多個點雙連通分量，需要防止重複計算 
              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++) //注意輸入時空格的影響，是以從1開始 
            {
                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;
}
           

題目（ 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.