private void solve(IDirectedGraph g)
{
result = new List <int>();
Degrees degrees_info = new Degrees(g);
int[] degrees = new int[g.V];
for (int i = 0; i < g.V; ++i)
{
degrees[i] = degrees_info.InDegree(i);
}
Queue <int> starts = new Queue <int>();
foreach (Int32 s in degrees_info.Sources())
{
starts.Enqueue(s);
}
//只要还有顶点
while (starts.Count > 0)
{
Int32 v = starts.Dequeue();
result.Add(v);
foreach (var w in g.Adj(v))
{
//删除这个顶点的指出的边，因为这个顶点已经被删除了
--degrees[w];
if (degrees[w] == 0)
{
//下一个拓扑排序的顺序
starts.Enqueue(w);
}
}
}
}
private void dfs(IDirectedGraph digraph, int v)
{
onStack[v] = true;
marked[v] = true;
foreach (int w in digraph.Adj(v))
{
if (HasCycle())
{
return;
}
else if (!marked[w])
{
edgeTo[w] = v;
dfs(digraph, w);
}
else if (onStack[w])//如果访问过且在一个路径中
{
cycle = new Stack <int>();
for (int x = v; x != w; x = edgeTo[x])
{
cycle.Push(x);//把前驱放入栈中直到环的起点的后继
}
cycle.Push(w);
cycle.Push(v);//为何从v开始呢？因为v->w是有连接的。此时栈顶和栈底都是一个
}
}
onStack[v] = false;//循环结束，不是一条栈上链的
}
private void dfs(IDirectedGraph g, int v)
{
marked[v] = true;
id[v] = count;
foreach (var w in g.Adj(v))
{
if (!marked[w])
{
dfs(g, w);
}
}
}
