如果在任意两个顶点之间存在两个不相交的顶点路径,则无向图被称为双向图。换句话说,我们可以说任意两个顶点之间都有一个循环。
可以说,如果图G是连通的,则它是一个双连通图,并且图中不存在关节点或切点。
为了解决这个问题,我们将使用DFS遍历。使用DFS,我们将尝试查找是否存在任何连接点。我们还检查DFS是否访问了所有顶点,如果不是,我们可以说该图未连接。
Input: The adjacency matrix of the graph. 0 1 1 1 0 1 0 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 1 1 0 Output: 该图是一个双向图。
isArticulation (开始,访问,光盘,低位,父级)
输入: 起始顶点,用来标记何时访问节点的访问数组,光盘将保留该顶点的发现时间,而低位将保留有关子树的信息。父级将保留当前顶点的父级。
输出-如果找到任何铰接点,则为True。
Begin time := 0 //the value of time will not be initialized for next function calls dfsChild := 0 mark start as visited set disc[start] := time+1 and low[start] := time + 1 time := time + 1 for all vertex v in the graph G, do if there is an edge between (start, v), then if v is visited, then increase dfsChild parent[v] := start if isArticulation(v, visited, disc, low, parent) is true, then return ture low[start] := minimum of low[start] and low[v] if parent[start] is φ AND dfsChild > 1, then return true if parent[start] is φ AND low[v] >= disc[start], then return true else if v is not the parent of start, then low[start] := minimum of low[start] and disc[v] done return false End
isBiconnected(图)
输入:给定的图形。
输出-如果图形是双向连接,则为True。
Begin initially set all vertices are unvisited and parent of each vertices are φ if isArticulation(0, visited, disc, low, parent) = true, then return false for each node i of the graph, do if i is not visited, then return false done return true End
#include<iostream> #define NODE 5 using namespace std; int graph[NODE][NODE] = { {0, 1, 1, 1, 0}, {1, 0, 1, 0, 0}, {1, 1, 0, 0, 0}, {1, 0, 0, 0, 1}, {0, 0, 0, 1, 0} }; int min(int a, int b) { return (a<b)?a:b; } bool isArticulation(int start, bool visited[], int disc[], int low[], int parent[]) { static int time = 0; int dfsChild = 0; visited[start] = true; //make the first vertex is visited disc[start] = low[start] = ++time; //initialize discovery time and the low time for(int v = 0; v<NODE; v++) { if(graph[start][v]) { //for all vertex v, which is connected with start if(!visited[v]) { dfsChild++; parent[v] = start; //make start node as parent if(isArticulation(v, visited, disc, low, parent)) return true; low[start] = min(low[start], low[v]); //when subtree from v is connected to one of parent of start node if(parent[start] == -1 && dfsChild > 1) { //when u have 2 or more children return true; } if(parent[start] != -1 && low[v]>= disc[start]) return true; } else if(v != parent[start]) //update low of start for previous call low[start] = min(low[start], disc[v]); } } return false; } bool isBiConnected() { bool *vis = new bool[NODE]; int *disc = new int[NODE]; int *low = new int[NODE]; int *parent = new int[NODE]; for(int i = 0; i<NODE; i++) { vis[i] = false; //no node is visited parent[i] = -1; //initially there is no parent for any node } if(isArticulation(0, vis, disc, low, parent)) //when no articulation point is found return false; for(int i = 0; i<NODE; i++) if(!vis[i]) //if any node is unvisited, the graph is not connected return false; return true; } int main() { if(isBiConnected()) cout << "该图是一个双向图。"; else cout << "该图不是双向图。"; }
输出结果
该图是一个双向图。