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.

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

55 1 2 3 4062 1 35 4 6 200

Sample Output

12

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.

Source

Central Europe 1996
求割点的满足条件：
1：若深度优先生成树的根有两棵或两棵以上的子树，则此根为割点。
2：若某非叶子结点v的孩子节点u没有指向v祖先的回边，则v点为割点。

#include <iostream>#include<string.h>using namespace std;#include<stdio.h>#include<vector>#include<algorithm>#define maxn 105vector<int>mat[105];int dfn[maxn],low[maxn],vis[maxn],root,n;int flag[maxn],scnt;void dfs(int v){    int i,w,cnt=0;//cnt表示孩子的个数。    scnt++;    vis[v]=1;    dfn[v]=low[v]=scnt;//记录时间戳    for(i=0;i<mat[v].size();i++)    {        w=mat[v][i];        if(!vis[w])//w没有访问过，则w是v的孩子节点        {            cnt++;            dfs(w);            low[v]=min(low[v],low[w]);            if(v==root&&cnt>1)//如果v是根并且孩子的个数大于1，则为割点                flag[v]=1;            if(v!=root&&low[w]>=dfn[v])//如果v不是根并且w及其子孙均无指向v的祖先的回边，则为割点。                flag[v]=1;        }        else if(v!=w)//w已访问过，说明v到w有回边。        {            low[v]=min(low[v],dfn[w]);        }    }}int main(){    int i,a,b,ans;    char c;  while(cin>>n)  {    if(n==0)break;    memset(flag,0,sizeof(flag));    memset(dfn,0,sizeof(dfn));    memset(low,0,sizeof(low));    memset(vis,0,sizeof(vis));    ans=scnt=0;root=2;    for(i=0;i<=n;i++)        mat[i].clear();   while(cin>>a)   {   if(a==0)break;     while((c=getchar())!='\n')     {         cin>>b;         mat[a].push_back(b);         mat[b].push_back(a);     }   }        dfs(root);     for(i=1;i<=n;i++)        if(flag[i])          ans++;        cout<<ans<<endl;  }    return 0;}