在此程序中,我们需要找到图的边缘连通性。图的图的边缘连通性意味着它是一个桥,将其删除将断开连接。随着断开的无向图中的桥的去除,连接组件的数量增加。
Begin Function connections() is a recursive function to find out the connections: A) Mark the current node un visited. B) Initialize time and low value C) Go through all vertices adjacent to this D) Check if the subtree rooted with x has a connection to one of the ancestors of w. 如果从x下的子树可到达的最低顶点是 below u in DFS tree, then w-x has a connection. E) Update low value of w for parent function calls. End
Con()
Begin Function Con() that uses connections(): A)将所有顶点标记为未访问. B) Initialize par and visited, and connections. C) Print the connections between the edges in the graph. End
#include<iostream> #include <list> #define N -1 using namespace std; class G { //功能声明 int n; list<int> *adj; void connections(int n, bool visited[], int disc[], int low[], int par[]); public: G(int n); //constructor void addEd(int w, int x); void Con(); }; 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::connections(int w, bool visited[], int dis[], int low[], int par[]) { static int t = 0; //将当前节点标记为已访问 visited[w] = true; dis[w] = low[w] = ++t; //遍历所有相邻的顶点 list<int>::iterator i; for (i = adj[w].begin(); i != adj[w].end(); ++i) { int x = *i; //x is current adjacent if (!visited[x]) { par[x] = w; connections(x, visited, dis, low, par); low[w] = min(low[w], low[x]); //如果从x下的子树可到达的最低顶点是 below w in DFS tree, then w-x is a connection if (low[x] > dis[w]) cout << w << " " << x << endl; } else if (x != par[w]) low[w] = min(low[w], dis[x]); } } void G::Con() { //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]; for (int i = 0; i < n; i++) { par[i] = N; visited[i] = false; } //call the function connections() to find edge connections for (int i = 0; i < n; i++) if (visited[i] == false) connections(i, visited, dis, low, par); } int main(){ cout << "\nConnections in first graph \n"; G g1(5); g1.addEd(1, 2); g1.addEd(3, 2); g1.addEd(2, 1); g1.addEd(0, 1); g1.addEd(1, 4); g1.Con(); return 0; }
输出结果
Connections in first graph 2 3 1 2 1 4 0 1