C ++程序查找图形中的铰接点数
图中的铰接点（或切点）是将其移除（以及通过它的边）会断开图的连接的点。断开的无向图的连接点是顶点移除，这增加了连接的组件数。
算法
Begin
We use dfs here to find articulation point:
In DFS, a vertex w is articulation point if one of the following two conditions is satisfied.
1) w is root of DFS tree and it has at least two children.
2) w is not root of DFS tree and it has a child x such that no vertex in subtree rooted with
w has a back edge to one of the ancestors of w in the tree.
End示例
#include<iostream>
#include <list>
#define N -1
using namespace std;
class G {
int n;
list<int> *adj;
//功能声明
void APT(int v, bool visited[], int dis[], int low[],
int par[], bool ap[]);
public:
G(int n); //constructor
void addEd(int w, int x);
void AP();
};
G::G(int n) {
this->n = n;
adj = new list<int>[n];
}
//在图上添加边
void G::addEd(int w, int x) {
adj[x].push_back(w); //add u to v's list
adj[w].push_back(x); //add v to u's list
}
void G::APT(int w, bool visited[], int dis[], int low[], int par[], bool ap[]) {
static int t=0;
int child = 0; //initialize child count of dfs tree is 0.
//将当前节点标记为已访问
visited[w] = true;
dis[w] = low[w] = ++t;
list<int>::iterator i;
//遍历所有相邻的顶点
for (i = adj[w].begin(); i != adj[w].end(); ++i) {
int x = *i; //x is current adjacent
if (!visited[x]) {
child++;
par[x] = w;
APT(x, visited, dis, low, par, ap);
low[w] = min(low[w], low[x]);
//w在以下情况下是一个关节点：
//w是DFS树的根，具有两个或多个孩子。
if (par[w] == N && child> 1)
ap[w] = true;
//如果w不是根，并且其子级之一的低值大于w的发现值。
if (par[w] != N && low[x] >= dis[w])
ap[w] = true;
} else if (x != par[w]) //update low value
low[w] = min(low[w], dis[x]);
}
}
void G::AP() {
//将所有顶点标记为未访问
bool *visited = new bool[n];
int *dis = new int[n];
int *low = new int[n];
int *par = new int[n];
bool *ap = new bool[n];
for (int i = 0; i < n; i++) {
par[i] = N;
visited[i] = false;
ap[i] = false;
}
// Call the APT() function to find articulation points in DFS tree rooted with vertex 'i'
for (int i = 0; i < n; i++)
if (visited[i] == false)
APT(i, visited, dis, low, par, ap);
//打印发音点
for (int i = 0; i < n; i++)
if (ap[i] == true)
cout << i << " ";
}
int main() {
cout << "\nArticulation points in first graph \n";
G g1(5);
g1.addEd(1, 2);
g1.addEd(3, 1);
g1.addEd(0, 2);
g1.addEd(2, 3);
g1.addEd(0, 4);
g1.AP();
return 0;
}输出结果
Articulation points in first graph 0 2