B

求割点的个数：
割点：在无向连通图中，如果将其中一个点及所连接该点的边去掉，图就不再连接，此点为割点。
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

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>using namespace std;const int MAXN = 20010;//点数const int MAXM = 50010;//边数struct Edge{    int to, next;} edge[MAXM];int head[MAXN], tot;int Low[MAXN], DFN[MAXN];int Index, top;int n, root, root_son, ans;int cut[MAXN];void addedge(int u, int v){    edge[tot].to = v;    edge[tot].next = head[u];    head[u] = tot++;}void Tarjan(int u){    Low[u] = DFN[u] = Index++;    for(int i=head[u]; i!=-1; i=edge[i].next)    {        int v = edge[i].to;        if(!DFN[v])        {            Tarjan(v);            if(u==root)                root_son++;//看看根节点有几条边            else            {                if(Low[u]>Low[v])                    Low[u] = Low[v];                if(Low[v]>=DFN[u])                    cut[u] = 1;//此点可以成为割点            }        }        else if(Low[u]>DFN[v])            Low[u] = DFN[v];    }}void init(){    tot = 0;//边数    memset(head, -1, sizeof(head));//邻接表的使用    memset(Low, 0, sizeof(Low));//追溯到最早的栈中的节点的次序号    memset(cut, 0, sizeof(cut));//当前的点是否可以成为割点    memset(DFN, 0, sizeof(DFN));//时间戳    Index = 1;//点数    root_son = ans = 0;    root = 1;//开始的根节点}int main(){    while(~scanf("%d", &n)&&n)    {        init();        int u, v;        while(~scanf("%d", &u)&&u)        {            while(getchar()!='\n')            {                scanf("%d", &v);                //无向图                addedge(u, v);                addedge(v, u);            }        }        Tarjan(root);        if(root_son>1)            cut[root] = 1;        for(int i=1; i<=n; i++)        {            if(cut[i])                ans++;        }        printf("%d\n", ans);    }    return 0;}