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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.
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.
The output contains for each block except the last in the input file one line containing the number of critical places.
5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0
1 2
题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=251
题意：有n个电话，有一些线连接在他们之间，有的电话如果不能工作了，则可能导致，n个电话不连通了，求出这样的电话又几个，其实就是求割点有多少个
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
#define maxn 10005
int dfn[maxn];///代表最先遍历到这个点的时间
int low[maxn];///这个点所能到达之前最早的时间点
int Father[maxn];///保存这个节点的父亲节点
int n, m, Time, top;///Time 时间点， top用于栈操作
vector<vector<int> > G;
void Init()
{
G.clear();
G.resize(n+1);
memset(low, 0, sizeof(low));
memset(dfn, 0, sizeof(dfn));
memset(Father, 0, sizeof(Father));
Time = 0;
}
void Tarjan(int u,int fa)
{
low[u] = dfn[u] = ++Time;
Father[u] = fa;
int len = G[u].size(), v;
for(int i=0; i<len; i++)
{
v = G[u][i];
if(!dfn[v])
{
Tarjan(v, u);
low[u] = min(low[u], low[v]);
}
else if(fa != v)///假如我们在这里写上了 low[u] = min(low[v], low[u]),那么就相当于我们由v回到了v之前的节点
low[u] = min(dfn[v], low[u]);
}
}
void solve()
{/**
求割点
一个顶点u是割点，当且仅当满足(1)或(2)
(1) u为树根，且u有多于一个子树。
(2) u不为树根，且满足存在(u,v)为树枝边(或称 父子边，即u为v在搜索树中的父亲)，使得 dfn(u)<=low(v)。
（也就是说 V 没办法绕过 u 点到达比 u dfn要小的点）
注：这里所说的树是指，DFS下的搜索树*/
int RootSon = 0, ans = 0;///根节点儿子的数量
bool Cut[maxn] = {false};///标记数组，判断这个点是否是割点
Tarjan(1,0);
for(int i=2; i<=n; i++)
{
int v = Father[i];
if(v == 1)///也是就说 i的父亲是根节点
RootSon ++;
else if(dfn[v] <= low[i])
Cut[v] = true;
}
for(int i=2; i<=n; i++)
{
if(Cut[i])
ans ++;
}
if(RootSon > 1)
ans++;
printf("%d\n", ans);
}
int main()
{
while(scanf("%d", &n), n)
{
int a, b;
char ch;
Init();
while(scanf("%d", &a), a)
{
while(scanf("%d%c",&b,&ch))
{
G[a].push_back(b);
G[b].push_back(a);
if(ch == ‘\n‘)
break;
}
}
solve();
}
return 0;
}
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原文地址：http://www.cnblogs.com/zhengguiping--9876/p/5498532.html
