本文整理汇总了C#中System.Collections.Queue.IsEmpty方法的典型用法代码示例。如果您正苦于以下问题：C# Queue.IsEmpty方法的具体用法？C# Queue.IsEmpty怎么用？C# Queue.IsEmpty使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类System.Collections.Queue的用法示例。
在下文中一共展示了Queue.IsEmpty方法的5个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: cycle
private IEnumerable<DirectedEdge> _cycle; // negative cycle (or null if no such cycle)
#endregion Fields
#region Constructors
/// <summary>
/// Computes a shortest paths tree from <tt>s</tt> to every other vertex in
/// the edge-weighted digraph <tt>G</tt>.
/// </summary>
/// <param name="g">g the acyclic digraph</param>
/// <param name="s">s the source vertex</param>
/// <exception cref="ArgumentException">unless 0 le; <tt>s</tt> le; <tt>V</tt> - 1</exception>
public BellmanFordSP(EdgeWeightedDigraph g, int s)
{
_distTo = new double[g.V];
_edgeTo = new DirectedEdge[g.V];
_onQueue = new bool[g.V];
for (var v = 0; v < g.V; v++)
_distTo[v] = double.PositiveInfinity;
_distTo[s] = 0.0;
// Bellman-Ford algorithm
_queue = new Collections.Queue<Integer>();
_queue.Enqueue(s);
_onQueue[s] = true;
while (!_queue.IsEmpty() && !HasNegativeCycle())
{
int v = _queue.Dequeue();
_onQueue[v] = false;
Relax(g, v);
}
}示例2: TopologicalX
private readonly int[] _rank; // rank[v] = order where vertex v appers in order
#endregion Fields
#region Constructors
/// <summary>
/// Determines whether the digraph <tt>G</tt> has a topological order and, if so,
/// finds such a topological order.
/// </summary>
/// <param name="g">g the digraph</param>
public TopologicalX(Digraph g)
{
// indegrees of remaining vertices
var indegree = new int[g.V];
for (var v = 0; v < g.V; v++)
{
indegree[v] = g.Indegree(v);
}
// initialize
_rank = new int[g.V];
_order = new Collections.Queue<Integer>();
var count = 0;
// initialize queue to contain all vertices with indegree = 0
var queue = new Collections.Queue<Integer>();
for (var v = 0; v < g.V; v++)
if (indegree[v] == 0) queue.Enqueue(v);
for (var j = 0; !queue.IsEmpty(); j++)
{
int v = queue.Dequeue();
_order.Enqueue(v);
_rank[v] = count++;
foreach (int w in g.Adj(v))
{
indegree[w]--;
if (indegree[w] == 0) queue.Enqueue(w);
}
}
// there is a directed cycle in subgraph of vertices with indegree >= 1.
if (count != g.V)
{
_order = null;
}
//assert check(G);
}示例3: Bfs
/// <summary>
/// breadth-first search from multiple sources
/// </summary>
/// <param name="g"></param>
/// <param name="sources"></param>
private void Bfs(Graph g, IEnumerable<Integer> sources)
{
var q = new Collections.Queue<Integer>();
foreach (int s in sources)
{
_marked[s] = true;
_distTo[s] = 0;
q.Enqueue(s);
}
while (!q.IsEmpty())
{
int v = q.Dequeue();
foreach (int w in g.Adj(v))
{
if (_marked[w]) continue;
_edgeTo[w] = v;
_distTo[w] = _distTo[v] + 1;
_marked[w] = true;
q.Enqueue(w);
}
}
}示例4: Bfs
private void Bfs(Graph g, int s)
{
var q = new Collections.Queue<Integer>();
_color[s] = WHITE;
_marked[s] = true;
q.Enqueue(s);
while (!q.IsEmpty())
{
int v = q.Dequeue();
foreach (int w in g.Adj(v))
{
if (!_marked[w])
{
_marked[w] = true;
_edgeTo[w] = v;
_color[w] = !_color[v];
q.Enqueue(w);
}
else if (_color[w] == _color[v])
{
_isBipartite = false;
// to form odd cycle, consider s-v path and s-w path
// and let x be closest node to v and w common to two paths
// then (w-x path) + (x-v path) + (edge v-w) is an odd-length cycle
// Note: distTo[v] == distTo[w];
_cycle = new Collections.Queue<Integer>();
var stack = new Collections.Stack<Integer>();
int x = v, y = w;
while (x != y)
{
stack.Push(x);
_cycle.Enqueue(y);
x = _edgeTo[x];
y = _edgeTo[y];
}
stack.Push(x);
while (!stack.IsEmpty())
_cycle.Enqueue(stack.Pop());
_cycle.Enqueue(w);
return;
}
}
}
}示例5: DirectedCycleX
private readonly Collections.Stack<Integer> _cycle; // the directed cycle; null if digraph is acyclic
#endregion Fields
#region Constructors
public DirectedCycleX(Digraph g)
{
// indegrees of remaining vertices
var indegree = new int[g.V];
for (var v = 0; v < g.V; v++)
{
indegree[v] = g.Indegree(v);
}
// initialize queue to contain all vertices with indegree = 0
var queue = new Collections.Queue<Integer>();
for (var v = 0; v < g.V; v++)
if (indegree[v] == 0) queue.Enqueue(v);
for (var j = 0; !queue.IsEmpty(); j++)
{
int v = queue.Dequeue();
foreach (int w in g.Adj(v))
{
indegree[w]--;
if (indegree[w] == 0) queue.Enqueue(w);
}
}
// there is a directed cycle in subgraph of vertices with indegree >= 1.
var edgeTo = new int[g.V];
var root = -1; // any vertex with indegree >= -1
for (var v = 0; v < g.V; v++)
{
if (indegree[v] == 0) continue;
root = v;
foreach (int w in g.Adj(v))
{
if (indegree[w] > 0)
{
edgeTo[w] = v;
}
}
}
if (root != -1)
{
// find any vertex on cycle
var visited = new bool[g.V];
while (!visited[root])
{
visited[root] = true;
root = edgeTo[root];
}
// extract cycle
_cycle = new Collections.Stack<Integer>();
var v = root;
do
{
_cycle.Push(v);
v = edgeTo[v];
} while (v != root);
_cycle.Push(root);
}
//assert check();
}