POJ 1144 Network

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.

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【题目分析】
求割点的一道裸题。很好写，也是用tarjan算法来做。真是不明白poj上怎么也全是这种坑爹的读入数据，害我想了半天。

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【代码】

#include <cstdio>
#include <cstring>
#define M(a) memset(a,-1,sizeof a)
#define M0(a) memset(a,0,sizeof a)
char ch;
int idx,x,y,n,rt,h[10001],ne[10001],en=0,ans,fr[10001],to[10001],dfn[10001],low[10001];
bool b[10001];
inline void add(int a,int b)
{fr[en]=a;to[en]=b;ne[en]=h[a];h[a]=en++;}
inline int min(int a,int b)
{return a>b?b:a;}
inline void tarjan(int u)
{
    int tot=0;
    low[u]=dfn[u]=++idx;
    for (int i=h[u];i>=0;i=ne[i])
    {
        if (!dfn[to[i]])
        {
            tot++,tarjan(to[i]),low[u]=min(low[u],low[to[i]]);
            if ((((u==rt)&&(tot>=2))||((u!=rt)&&(low[to[i]]>=dfn[u])))&&b[u]==false) b[u]=true,ans++;
        }
        else low[u]=min(dfn[to[i]],low[u]);
    }
}
int main()
{
    while (scanf("%d",&n)&&n)
    {
        memset(b,false,sizeof b);
        idx=0;M(h);M(ne);M(fr);M(to);en=0;M0(dfn);M0(low);ans=0;
        while (scanf("%d",&x)&&x)
            while (scanf("%c",&ch)&&ch==' ')
                scanf("%d",&y),add(x,y),add(y,x);
        rt=1,tarjan(1);
        printf("%d\n",ans);
    }
}