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.
割点：去除这个点之后连通分量增加
割边：去除这条边之后连通分量增加
DFS判断割点的方法：
①s是根节点且有两个或更多的子节点
②非根节点u，当且仅当u存在一个子节点v，v及其后代都没有回退边连回u的祖先
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int N=109;
int low[N],num[N],dfn; //num[u]记录DFS每个点的访问顺序 low[u]记录v和v的后代能连回到祖先的num
bool iscut[N];
vector <int> G[N];
void dfs(int u, int fa){ //u的父结点是fa
low[u] = num[u] = ++ dfn; //初始值
int child = 0; //孩子数目
for (int i = 0;i < G[u].size(); i++) { //处理u的所有子结点
int v = G[u][i];
if (!num[v]) { //v没访问过
child++;
dfs(v, u);
low[u] = min(low[v], low[u]); //用后代的返回值更新low值
if (low[v] >= num[u] && u !=1)
iscut[u] = true; //标记割点 //if(low[v]>num[u]&&u!=1) iscut[u]=true 判断割边
}
else if(num[v] < num[u] && v != fa)
//处理回退边，注意这里v != fa，fa是u的父结点，
//fa也是u的邻居，但是前面已经访问过，不需要处理它
low[u] = min(low[u], num[v]);
}
if (u == 1 && child >= 2) //根结点，有两个以上不相连的子树
iscut[1] = true;
}
int main(){
int ans,n;
while (scanf("%d",&n)!=-1){
if (n==0)break;
memset(low,0,sizeof(low));
memset(num,0,sizeof(num));
dfn=0;
for (int i=0;i<=n;i++) G[i].clear();
int a,b;
while (scanf("%d",&a)&&a)
while (getchar()!=‘\n‘){
scanf("%d",&b);
G[a].push_back(b);
G[b].push_back(a);
}
memset(iscut,false,sizeof(iscut));
ans = 0;
dfs(1,1);
for (int i=1;i<=n;i++) ans+=iscut[i];
printf("%d\n",ans);
}
}
原文地址：https://www.cnblogs.com/lyj1/p/11392798.html