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法律研究使用 DFS 检查有向图是否为树的 C++ 程序
如果图不包含任何循环，则该图是一棵树。这是一个使用 DFS 检查有向图是否为树的 C++ 程序。
算法
Begin function cyclicUtil() : a) Mark the current node as visited and part of recursion stack b) Recur for all the vertices adjacent to this vertex. c) Remove the vertex from recursion stack. function cyclic() : a) Mark all the vertices as not visited and not part of recursion stack b) Call the CyclicUtill() function to detect cycle in different trees End
示例
#include<iostream>
#include <list>
#include <limits.h>
using namespace std;
class G {
int n;
list<int> *adj; //contain adjacent list.
bool CyclicUtil(int v, bool visited[], bool *rs);
public:
G(int V); // Constructor
void addEd(int v, int w);
bool cyclic();
};
G::G(int n) {
this->n = n;
adj = new list<int> [n];
}
void G::addEd(int v, int u) //to add edges in the graph {
adj[v].push_back(u); //add u to v’s list
}
bool G::CyclicUtil(int v, bool visited[], bool *recurS) {
if (visited[v] == false) {
visited[v] = true; //Mark the current node as visited and part of recursion stack
recurS[v] = true;
// Recur for all the vertices adjacent to this vertex.
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i) {
if (!visited[*i] && CyclicUtil(*i, visited, recurS))
return true;
else if (recurS[*i])
return true;
}
}
recurS[v] = false; //Remove the vertex from recursion stack.
return false;
}
//check if the graph is tree or not
bool G::cyclic() {
//Mark all the vertices as not visited and not part of recursion stack
bool *visited = new bool[n];
bool *recurS = new bool[n];
for (int i = 0; i < n; i++) {
visited[i] = false;
recurS[i] = false;
}
// Call the CyclicUtill() function to detect cycle in different trees
for (int i = 0; i < n; i++)
if (CyclicUtil(i, visited, recurS))
return true;
return false;
}
int main() {
G g(4);
g.addEd(0, 2);
g.addEd(1, 2);
g.addEd(2, 0);
g.addEd(3, 2);
if (g.cyclic())
cout << "Directed Graph isn't a tree";
else
cout << "Directed Graph is a tree";
return 0;
}输出
Directed Graph isn't a tree
更新日期： 30-Jul-2019
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