POJ -1144 Network 求割点

                                      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.

点u是割点的充分条件：

1. 如果u是dfs树的根，则u至少有两个子女。

2. 如果u不是dfs树的根，则它至少有一个子女w，从w出发，不可能通过和w、w的子孙相连的回边到达u的祖先。

u是割点的充要条件是：u或则是具有两个以上子女的深度优先生成树的根，或则虽然不是根，但它有一个子女w，使得low[w]>=dfn[u]。

dfn数组是记录时间 从根出发dfs遍历树的节点的时间。 第二个条件大概意思就是u的子女必须通过u到达u的祖先(接近根的节点

代码如下：

   
/* 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])//这里的意思 就是v点已经访问了 v点是u点的祖先 及v到u有回边
        {
           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);
            }
        }//这里懵了好久 输入有点坑
        times=0;
        roots=1;
        dfs(roots);
        for(i=1; i<=n; i++)
            if(dis[i])ans++;
        printf("%d\n",ans);
    }
    return 0;
}