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我有类似的问题 - 相同 - 与此 one .所以我想知道如何不仅检测循环而且打印出这个循环包含的顶点。我尝试了上面问题中提到的方法,但我一定做错了什么,为什么它对我不起作用。我的程序还检查一个特定的顶点是否形成循环。我的代码在这里:
#include<iostream>
#include <list>
using namespace std;
class Graph
{
int V;
list<int> *adj;
public:
Graph(int V);
void addEdge(int v, int w);
bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
bool Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent, int index)
{
visited[v] = true;
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
{
if (!visited[*i])
{
if (isCyclicUtil(*i, visited, cycleVertices, v, index)) {
if (index <= 1 || cycleVertices[0] != cycleVertices[index - 1])
cycleVertices[index++] = *i;
return true;
}
}
else if (*i != parent) {
cycleVertices[index++] = *i;
return true;
}
}
return false;
}
int main()
{
bool *visited = new bool[5];
for (int i = 0; i < 5; i++)
visited[i] = false;
int cycleVertices[5];
for (int i = 0; i < 5; i++)
cycleVertices[i] = -1;
Graph g1(5);
g1.addEdge(1, 0);
g1.addEdge(0, 2);
g1.addEdge(2, 1);
g1.addEdge(0, 3);
g1.addEdge(3, 4);
g1.isCyclicUtil(4, visited, cycleVertices, -1, 0) ? cout << "Graph contains cycle\n" :
cout << "Graph doesn't contain cycle\n";
int x = 0;
while (cycleVertices[x] != -1)
cout << cycleVertices[x++] << " ";
return 0;
}
最佳答案
我找到了解决方案。我在这个 post 中尝试了 j_random_hacker 的解决方案它没有用。但问题在于我的代码中的 cycleVertices 中的索引。变量索引总是相同的。所以我在类 Graph 中添加了一个新的属性索引,现在它可以工作了。所以这是编辑后的代码:
#include<iostream>
#include <list>
#define FINISHED -1
#define NOCYCLE -2
using namespace std;
class Graph
{
int V;
int index;
list<int> *adj;
public:
Graph(int V);
void addEdge(int v, int w);
void set_index();
int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent);
};
Graph::Graph(int V)
{
this->V = V;
adj = new list<int>[V];
this->index = 0;
}
void Graph::set_index()
{
this->index = 0;
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w);
adj[w].push_back(v);
}
int Graph::isCyclicUtil(int v, bool visited[], int *cycleVertices, int parent)
{
visited[v] = true;
list<int>::iterator i;
for (i = adj[v].begin(); i != adj[v].end(); ++i)
{
if (!visited[*i])
{
int result = isCyclicUtil(*i, visited, cycleVertices, v);
if (result == FINISHED)
return FINISHED;
else if (result != NOCYCLE) {
cycleVertices[index++] = v;
if (result == v)
return FINISHED;
else
return result;
}
}
else if (*i != parent) {
return *i;
}
}
return NOCYCLE;
}
int main()
{
bool *visited = new bool[4];
for (int i = 0; i < 4; i++)
visited[i] = false;
int cycleVertices[4];
for (int i = 0; i < 4; i++)
cycleVertices[i] = -1;
Graph g1(4);
g1.addEdge(0, 1);
g1.addEdge(1, 2);
g1.addEdge(2, 3);
g1.addEdge(3, 0);
g1.isCyclicUtil(3, visited, cycleVertices, -1) ? cout << "Graph contains cycle\n" :
cout << "Graph doesn't contain cycle\n";
int x = 0;
while (cycleVertices[x] != -1)
cout << cycleVertices[x++] << " ";
return 0;
}
关于c++ - 检测和打印无向图中的循环,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32873133/
为了获得我的瓷砖,我这样做: style(styleUri = Style.MAPBOX_STREETS) { +vectorSource(id = "parcel-source") {
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