POJ1144——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

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.

Source

Central Europe 1996

求某些点，这些点出问题了会牵涉到其他的点，其实这些点就是割点，去掉这些点，整张图的连通块就多了。

简单介绍下tarjan算法求割点的原理：
  假设现在递归到点u
  1.u是dfs搜索树的根，则u必须有2棵以上的子树，u就是割点
  2.u不是根，但是对于边(u,v)的点，有low[v]>=DFN[u]，那么u就是割点

1的话，很显然，如果u是根，去掉这个点，它的那些子树一定不连通，那么就是割点了
2.回想下，low[]和DFN[]的意义，如果说low[v]<DFN[u],那么说明从v开始有其他边到它的祖先，所以就算去掉了点u，连通块还是不会增加，相反，如果v回到它的祖先必须经过u，那么去掉u之后，连通块就多了。

#include<string.h>
#include<stdio.h>

const int maxn=150;
int n;
struct node
{
	int to;
	int next;
}edge[maxn*maxn];
int head[maxn];
int index;
int tot;
int low[maxn];
int DFN[maxn];
bool instack[maxn];
bool cut_point[maxn];

void init()
{
	tot=0;
	index=0;
	memset(head,-1,sizeof(head));
	memset(cut_point,0,sizeof(cut_point));
	memset(instack,0,sizeof(instack));
}

void addedge(int from,int to)
{
	edge[tot].to=to;
	edge[tot].next=head[from];
	head[from]=tot++;
}

void tarjan(int u,int root,int fa)
{
	DFN[u]=low[u]=++index;
	instack[u]=1;
	int cnt=0;
	for(int i=head[u];i!=-1;i=edge[i].next)
	{
		int v=edge[i].to;
		if(!instack[v])
		{
			tarjan(v,root,u);
			cnt++;
			if(low[u]>low[v])
		 		low[u]=low[v];
 			if(u==root && cnt>1)
 				cut_point[u]=1;
			else if(u!=root && low[v]>=DFN[u])
				cut_point[u]=1;
		}
		else if(v!=fa && low[u] > DFN[v])
			low[u]=DFN[v];
	}
}

int search_cut_point()
{
	int ans=0;
	tarjan(1,1,-1);
	for(int i=1;i<=n;i++)
		if(cut_point[i])
			ans++;
	return ans;
}

int main()
{
	while(~scanf("%d",&n) && n)
	{
		init();
		int s,t;
		while(scanf("%d",&s) && s)
		{
			while(getchar()!='\n')
			{
				scanf("%d",&t);
				addedge(s,t);
				addedge(t,s);
			}	
		}
		printf("%d\n",search_cut_point());
	}
	return 0;
}