POJ -1144 Network 求割點

                                      Network

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.

點u是割點的充分條件：

1. 如果u是dfs樹的根，則u至少有兩個子女。

2. 如果u不是dfs樹的根，則它至少有一個子女w，從w出發，不可能通過和w、w的子孫相連的回邊到達u的祖先。

u是割點的充要條件是：u或則是具有兩個以上子女的深度優先生成樹的根，或則雖然不是根，但它有一個子女w，使得low[w]>=dfn[u]。

dfn數組是記錄時間 從根出發dfs遍歷樹的節點的時間。 第二個條件大概意思就是u的子女必須通過u到達u的祖先(接近根的節點

代碼如下：

   
/* POJ 1144*/


#include<iostream>
#include<cstdio>
#include<cstring>
#include<ctime>
#include<cstdlib>
#include<cmath>
#include<string>
#include<queue>
#include<vector>
#include<stack>
#include<list>
#include<set>
#define manx maxn
const int maxn = 100 +10;
using namespace std;
int times,roots,cnt;
int Next[maxn*200],To[maxn*200],Lext[maxn],vis[manx],dis[manx];
int dfn[manx],low[manx];
void init()
{
    times=roots=cnt=0;
    memset(Next,0,sizeof(Next));
    memset(Lext,0,sizeof(Lext));
    memset(To,0,sizeof(To));
    memset(vis,0,sizeof(vis));
    memset(dis,0,sizeof(dis));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
}
void add(int u,int v)
{
    Next[++cnt]=Lext[u];
    To[cnt]=v;
    Lext[u]=cnt;
}
int dfs(int u)
{
    int cntn=0;
    dfn[u]=low[u]=++times;
    vis[u]=1;
    for(int i=Lext[u]; i; i=Next[i])
    {
        int v=To[i];
        if(!dfn[v])
        {
            cntn++;
            dfs(v);
            low[u]=min(low[u],low[v]);
            if(u==roots&&cntn==2)
                dis[u]=1;
            if(u!=roots&&low[v]>=dfn[u])
                dis[u]=1;
        }
        else if(vis[v])//這裏的意思 就是v點已經訪問了 v點是u點的祖先 及v到u有回邊
        {
           low[u]=min(low[u],dfn[v]);
        }
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)==1&&n)
    {
        int a,b;
        int i,ans=0;
        char c;
        init();
        while(scanf("%d",&a)==1&&a)
        {
            while((c=getchar())!='\n')
            {
                scanf("%d",&b);
                add(a,b);
                add(b,a);
            }
        }//這裏懵了好久 輸入有點坑
        times=0;
        roots=1;
        dfs(roots);
        for(i=1; i<=n; i++)
            if(dis[i])ans++;
        printf("%d\n",ans);
    }
    return 0;
}