poj 1144 Network tarjan求无向连通图的割点个数

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.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=100000;
//tarjan算法求无向连通图的割点个数
struct edge
{
int t,w;
int next;
};
int V;//定点个数
int p[maxn];
edge G[maxn];
int l;
void init()
{
memset(p,-1,sizeof(p));
l=0;
}
void addedge(int u,int t,int w,int l)
{
G[l].w=w;
G[l].t=t;
G[l].next=p[u];
p[u]=l;
}
//tarjan 求割点 割边
int cut[maxn];//cut[i]非0表示i是割点
int color[maxn];//颜色:0表示没有访问,1表示正在访问,2表示访问结束
int lowc[maxn];//表示i及i的子孙相连的辈分最高的祖先节点所在的深度
int d[maxn];//表示i节点在树中的深度
int root;//根节点
int fath;//父节点
int pcnt;//割点个数
int egcnt;//割边个数
void dfs(int u,int fath,int deep)
{
    color[u]=1;//正在访问
    lowc[u]=d[u]=deep;//深度
    int tot=0;//子树个数
    for(int i=p[u];i!=-1;i=G[i].next)
    {
        int t=G[i].t;
        if(t!=fath&&color[t]==1)
        {
            lowc[u]=min(lowc[u],d[t]);
        }
        if(color[t]==0)
        {
            dfs(t,u,deep+1);
            tot++;//子树加1
            lowc[u]=min(lowc[u],lowc[t]);
            //求割点
            if((u==root&&tot>1)||(u!=root&&lowc[t]>=d[u])) cut[u]=1;//不能将pscnt++写到这里
            //求割边
            //if(lowc[t]>d[u]) edge[u][t]=true;  u->t是割边
        }
    }
    color[u]=2;
}
void calc()
{
    pcnt=egcnt=0;
    memset(cut,0,sizeof(cut));
    memset(color,0,sizeof(color));
    memset(lowc,0,sizeof(lowc));
    memset(d,0,sizeof(d));
    root=1;
    dfs(root,-1,1);
    for(int i=1;i<=V;i++) if(cut[i]) pcnt++;
}
int main()
{
while(scanf("%d",&V)&&V)
    {
    init();//初始化
        int id;
        while(scanf("%d",&id)&&id)
{
while(getchar() != '\n')
            {
                int tmp; scanf("%d", &tmp);
                //插入无向边
                addedge(id, tmp,1,l++), addedge(tmp, id,1,l++);
            }
}
calc();//求割点
printf("%d\n",pcnt);
}
return 0;
}