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 adirect connection, it can go through several exchanges. From time to time thepower supply fails at a place and then the exchange does not operate. Theofficials from TLC realized that in such a case it can happen that besides thefact that the place with the failure is unreachable, this can also cause thatsome other places cannot connect to each other. In such a case we will say theplace (where the failure 
occured) is critical. Now the officials are trying to write a program forfinding the number of all such critical places. Help them.

Input

The input file consists of several blocks of lines. Each block describesone 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 placefollowed by the numbers of some places to which there is a direct line fromthis place. These at most N lines completely describe the network, i.e., eachdirect connection of two places in the network is contained at least in onerow. All numbers in one line are separated 
by one space. Each block ends with a line containing just 0. The last block hasonly one line with N = 0;

Output

The output contains for each block except the last in the input file oneline 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 todetermine,there are no extra blank before the end of each line.

Source

Central Europe1996

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define N 105using namespace std;struct node {	int v,w,next;} edge[N*N];int dfn[N],low[N],vis[N],count[N],head[N];int n,m,num;void addedge(int u,int v) {	edge[num].v=v;	edge[num].next=head[u];	head[u]=num++;}void tarjan(int i) {	int j,k;	low[i]=dfn[i]=m++;	vis[i]=1;	for(j=head[i]; j!=-1; j=edge[j].next) {		k=edge[j].v;		if(!vis[k]) {			tarjan(k);			low[i]=min(low[k],low[i]);			if(low[k]>=dfn[i])				count[i]++;		} else			low[i]=min(low[i],dfn[k]);	}}int trans(char *s) {	int i,j;	i=j=0;	while(s[j]) {		i=i*10+s[j]-‘0‘;		j++;	}	return i;}int main() {	char st[10],ch;	int i,j,ans;	while(1) {		scanf("%d",&n);		memset(head,-1,sizeof(head));		memset(count,0,sizeof(count));		memset(vis,0,sizeof(vis));		m=num=ans=0;		if(n==0)			break;		getchar();		while(1) {			scanf("%s",st);			if(st[0]==‘0‘)				break;			i=trans(st);			while(1) {				ch=getchar();				if(ch==‘\n‘)					break;				scanf("%s",st);				j=trans(st);				addedge(i,j);				addedge(j,i);			}		}		tarjan(1);		for(i=2; i<=n; i++)			if(count[i])				ans++;		if(count[1]>1)			ans++;		printf("%d\n",ans);	}	return 0;}

思路：就是求割点的个数，还有就是读入的各种处理很麻烦。

POJ 1144 Network