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我正在解决一个优化问题，除其他外，我必须最大化流网络。我实现了一个基于 C++ 代码的流最大化算法，该算法基于Sedgewick “Java 中的算法，第三版，第 5 部分：图形算法”一书中出现的以下java 代码，该算法使用基于顶点的 PREFLOW- 最大化网络流-推送算法：

class NetworkMaxFlow
{ private Network G; private int s, t;
  private int[] h, wt;
  private void initheights()
  NetworkMaxFlow(Network G, int s, int t)
  { this.G = G; this.s = s; this.t = t;
    wt = new int[G.V()]; h = new int[G.V()];
    initheights();
    intGQ gQ = new intGQ(G.V());
    gQ.put(s); wt[t] = -(wt[s] = Edge.M*G.V());
    while (!gQ.empty())
    { int v = gQ.get();
      AdjList A = G.getAdjList(v);
      for (Edge e = A.beg(); !A.end(); e = A.nxt())
        { int w = e.other(v), cap = e.capRto(w);
          int P = cap < wt[v] ? cap : wt[v];
          if (P > 0 && v == s || h[v] == h[w]+1) // first observation (see below)
            { e.addflowRto(w, P);
              wt[v] -= P; wt[w] += P;
              if ((w != s) && (w != t)) gQ.put(w); // enqueue w if it is not source or sink
            }
        }
      if (v != s && v != t && wt[v] > 0) // why check v != t if t never enter the queue?
        { h[v]++; gQ.put(v); }
    }
  }
}

我的实现，基于那段代码，无法最大化下面的网络 容量网络； 所有流量都为零 执行后，产生的流程如下 有了这个流程，流量值是8，但是最大值是9，如下图的 流程前一个网络的最终状态 所示 最大流量据我了解，该算法与书中的解释是一致的。但是，我看到了两件奇怪的事情

  1. 源中没有明确的预流阶段。当谓词为真时，它被包含在while并首先执行一次。P > 0 && v == s也许这样做是为了缩短代码
  2. 根据我的理解和书中的论述，sink 永远不会进入队列。但是，当高度增加时，代码会检查 v != t。这有什么原因吗？

这是我在 C++ 中实现该算法的摘录

template <class Net, class Q_Type> typename Net::Flow_Type
generic_preflow_vertex_push_maximum_flow(Net & net)
{
  init_height_in_nodes(net); // breadth first traverse from sink to
                 // source. Nodes are labeled with their
                 // minimal distance (in nodes) to sink
  auto source = net.get_source();
  auto sink   = net.get_sink();

  using Itor = __Net_Iterator<Net>;
  Q_Type q; // generic queue (can be fifo, heap or random) of active nodes

  // preflow: floods all nodes connected to the source 
  for (Itor it(source); it.has_curr(); it.next()) 
    {
      auto arc  = it.get_curr();  
      arc->flow = arc->cap; // saturate arc to its maximum 
      auto tgt = net.get_tgt_node(arc);
      put_in_active_queue(q, tgt);
      assert(node_height<Net>(source) == node_height<Net>(tgt) + 1);
      assert(not is_residual<Net>(source, arc));
    }

  while (not q.is_empty()) // while there are active nodes
    {
      auto src = get_from_active_queue(q);
      auto excess = net.get_in_flow(src) - net.get_out_flow(src);

      for (Itor it(src); it.has_curr(); it.next()) 
        {
          auto arc = it.get_curr();
          auto tgt = net.get_connected_node(arc, src);

          if (node_height<Net>(src) != node_height<Net>(tgt) + 1)
            continue; // this arc is not eligible

          typename Net::Flow_Type flow_to_push;
          if (is_residual<Net>(src, arc))
            {
              flow_to_push = std::min(arc->flow, excess);
              arc->flow -= flow_to_push;
            }
          else
            {
              flow_to_push = std::min(arc->cap - arc->flow, excess);
              arc->flow += flow_to_push;
            }

          excess -= flow_to_push;
          if (tgt != sink and tgt != source)
            put_in_active_queue(q, tgt);
        }

    if (excess > 0) // src still active?
      { 
        node_height<Net>(src)++;
        put_in_active_queue(q, src);
      }
  }

  return net.flow_value(); // sum of all outing flow from source
}

¿ 有人发现我的代码和 Sedgewick 的代码之间存在逻辑上的不一致？我的印象是我的代码（也许还有 Sedgewick）没有正确处理高度的增加。但我不明白为什么

我展示了无法最大化的网络的详细执行跟踪（跟踪从 while 的第一个 q.get() 开始。括号中的值是高度的值。IN 是到节点的传入流。OUT即将到来的一个。

例如，该行

    4104 (2) --> 0 (1) pushing 1 from 4104 toward 0

指符合条件的弧 4104-->0。节点4104的高度为2，节点0的高度为1。表述“推1”意味着1个单位的流被推向目标节点(0)。该行================分隔每个队列提取。队列是先进先出的，其状态在每次处理结束时打印。

请注意，多次推送或减少零流量单元，但目标节点变为活动状态。

这是执行跟踪

Initial Queue = 4104 4105 4106 4107 4108

Active node 4104 Height = 2 IN = 1 OUT = 0
    4104 (2) --> source (3) not eligible
    4104 (2) --> 0 (1) pushing 1 from 4104 toward 0
    4104 (2) --> 1 (1) pushing 0 from 4104 toward 1
    4104 (2) --> 2 (1) pushing 0 from 4104 toward 2
    4104 (2) --> 4 (1) pushing 0 from 4104 toward 4
    Excess = 0
    Queue = 4105 4106 4107 4108 0 1 2 4
================
Active node 4105 Height = 2 IN = 3 OUT = 0
    4105 (2) --> source (3) not eligible
    4105 (2) --> 1 (1) pushing 1 from 4105 toward 1
    4105 (2) --> 4 (1) pushing 1 from 4105 toward 4
    4105 (2) --> 6 (1) pushing 1 from 4105 toward 6
    Excess = 0
    Queue = 4106 4107 4108 0 1 2 4 6
================
Active node 4106 Height = 2 IN = 1 OUT = 0
    4106 (2) --> source (3) not eligible
    4106 (2) --> 1 (1) pushing 1 from 4106 toward 1
    4106 (2) --> 5 (1) pushing 0 from 4106 toward 5
    Excess = 0
    Queue = 4107 4108 0 1 2 4 6 5
================
Active node 4107 Height = 2 IN = 1 OUT = 0
    4107 (2) --> source (3) not eligible
    4107 (2) --> 1 (1) pushing 1 from 4107 toward 1
    4107 (2) --> 2 (1) pushing 0 from 4107 toward 2
    4107 (2) --> 3 (1) pushing 0 from 4107 toward 3
    4107 (2) --> 4 (1) pushing 0 from 4107 toward 4
    4107 (2) --> 6 (1) pushing 0 from 4107 toward 6
    Excess = 0
    Queue = 4108 0 1 2 4 6 5 3
================
Active node 4108 Height = 2 IN = 3 OUT = 0
    4108 (2) --> source (3) not eligible
    4108 (2) --> 1 (1) pushing 1 from 4108 toward 1
    4108 (2) --> 2 (1) pushing 1 from 4108 toward 2
    4108 (2) --> 4 (1) pushing 1 from 4108 toward 4
    4108 (2) --> 5 (1) pushing 0 from 4108 toward 5
    4108 (2) --> 6 (1) pushing 0 from 4108 toward 6
    Excess = 0
    Queue = 0 1 2 4 6 5 3
================
Active node 0 Height = 1 IN = 1 OUT = 0
    0 (1) --> sink (0) pushing 1 from 0 toward sink
    0 (1) --> 4104 (2) not eligible
    Excess = 0
    Queue = 1 2 4 6 5 3
================
Active node 1 Height = 1 IN = 4 OUT = 0
    1 (1) --> sink (0) pushing 2 from 1 toward sink
    1 (1) --> 4105 (2) not eligible
    1 (1) --> 4106 (2) not eligible
    1 (1) --> 4107 (2) not eligible
    1 (1) --> 4108 (2) not eligible
    Excess = 2    1 goes back onto queue with label 2
    Queue = 2 4 6 5 3 1
================
Active node 2 Height = 1 IN = 1 OUT = 0
    2 (1) --> sink (0) pushing 1 from 2 toward sink
    2 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 4 6 5 3 1
================
Active node 4 Height = 1 IN = 2 OUT = 0
    4 (1) --> sink (0) pushing 2 from 4 toward sink
    4 (1) --> 4105 (2) not eligible
    4 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 6 5 3 1
================
Active node 6 Height = 1 IN = 1 OUT = 0
    6 (1) --> sink (0) pushing 1 from 6 toward sink
    6 (1) --> 4105 (2) not eligible
    Excess = 0
    Queue = 5 3 1
================
Active node 5 Height = 1 IN = 0 OUT = 0
    5 (1) --> sink (0) pushing 0 from 5 toward sink
    Excess = 0
    Queue = 3 1
================
Active node 3 Height = 1 IN = 0 OUT = 0
    3 (1) --> sink (0) pushing 0 from 3 toward sink
    Excess = 0
    Queue = 1
================
Active node 1 Height = 2 IN = 4 OUT = 2
    1 (2) --> 4105 (2) not eligible
    1 (2) --> 4106 (2) not eligible
    1 (2) --> 4107 (2) not eligible
    1 (2) --> 4108 (2) not eligible
    Excess = 2    1 goes back onto queue with label 3
    Queue = 1
================
Active node 1 Height = 3 IN = 4 OUT = 2
    1 (3) --> 4105 (2) Reducing 1 from 1 toward 4105
    1 (3) --> 4106 (2) Reducing 1 from 1 toward 4106
    1 (3) --> 4107 (2) Reducing 0 from 1 toward 4107
    1 (3) --> 4108 (2) Reducing 0 from 1 toward 4108
    Excess = 0
    Queue = 4105 4106 4107 4108
================
Active node 4105 Height = 2 IN = 3 OUT = 2
    4105 (2) --> source (3) not eligible
    4105 (2) --> 1 (3) not eligible
    Excess = 1    4105 goes back onto queue with label 3
    Queue = 4106 4107 4108 4105
================
Active node 4106 Height = 2 IN = 1 OUT = 0
    4106 (2) --> source (3) not eligible
    4106 (2) --> 1 (3) not eligible
    4106 (2) --> 5 (1) pushing 1 from 4106 toward 5
    Excess = 0
    Queue = 4107 4108 4105 5
================
Active node 4107 Height = 2 IN = 1 OUT = 1
    4107 (2) --> source (3) not eligible
    4107 (2) --> 2 (1) pushing 0 from 4107 toward 2
    4107 (2) --> 3 (1) pushing 0 from 4107 toward 3
    4107 (2) --> 4 (1) pushing 0 from 4107 toward 4
    4107 (2) --> 6 (1) pushing 0 from 4107 toward 6
    Excess = 0
    Queue = 4108 4105 5 2 3 4 6
================
Active node 4108 Height = 2 IN = 3 OUT = 3
    4108 (2) --> source (3) not eligible
    4108 (2) --> 5 (1) pushing 0 from 4108 toward 5
    4108 (2) --> 6 (1) pushing 0 from 4108 toward 6
    Excess = 0
    Queue = 4105 5 2 3 4 6
================
Active node 4105 Height = 3 IN = 3 OUT = 2
    4105 (3) --> source (3) not eligible
    4105 (3) --> 1 (3) not eligible
    Excess = 1    4105 goes back onto queue with label 4
    Queue = 5 2 3 4 6 4105
================
Active node 5 Height = 1 IN = 1 OUT = 0
    5 (1) --> sink (0) pushing 1 from 5 toward sink
    5 (1) --> 4106 (2) not eligible
    Excess = 0
    Queue = 2 3 4 6 4105
================
Active node 2 Height = 1 IN = 1 OUT = 1
    2 (1) --> sink (0) pushing 0 from 2 toward sink
    2 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 3 4 6 4105
================
Active node 3 Height = 1 IN = 0 OUT = 0
    3 (1) --> sink (0) pushing 0 from 3 toward sink
    Excess = 0
    Queue = 4 6 4105
================
Active node 4 Height = 1 IN = 2 OUT = 2
    4 (1) --> 4105 (4) not eligible
    4 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 6 4105
================
Active node 6 Height = 1 IN = 1 OUT = 1
    6 (1) --> sink (0) pushing 0 from 6 toward sink
    6 (1) --> 4105 (4) not eligible
    Excess = 0
    Queue = 4105
================
Active node 4105 Height = 4 IN = 3 OUT = 2
    4105 (4) --> source (3) Reducing 1 from 4105 toward source
    4105 (4) --> 1 (3) pushing 0 from 4105 toward 1
    Excess = 0
    Queue = 1
================
Active node 1 Height = 3 IN = 2 OUT = 2
    1 (3) --> 4107 (2) Reducing 0 from 1 toward 4107
    1 (3) --> 4108 (2) Reducing 0 from 1 toward 4108
    Excess = 0
    Queue = 4107 4108
================
Active node 4107 Height = 2 IN = 1 OUT = 1
    4107 (2) --> source (3) not eligible
    4107 (2) --> 2 (1) pushing 0 from 4107 toward 2
    4107 (2) --> 3 (1) pushing 0 from 4107 toward 3
    4107 (2) --> 4 (1) pushing 0 from 4107 toward 4
    4107 (2) --> 6 (1) pushing 0 from 4107 toward 6
    Excess = 0
    Queue = 4108 2 3 4 6
================
Active node 4108 Height = 2 IN = 3 OUT = 3
    4108 (2) --> source (3) not eligible
    4108 (2) --> 5 (1) pushing 0 from 4108 toward 5
    4108 (2) --> 6 (1) pushing 0 from 4108 toward 6
    Excess = 0
    Queue = 2 3 4 6 5
================
Active node 2 Height = 1 IN = 1 OUT = 1
    2 (1) --> sink (0) pushing 0 from 2 toward sink
    2 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 3 4 6 5
================
Active node 3 Height = 1 IN = 0 OUT = 0
    3 (1) --> sink (0) pushing 0 from 3 toward sink
    Excess = 0
    Queue = 4 6 5
================
Active node 4 Height = 1 IN = 2 OUT = 2
    4 (1) --> 4105 (4) not eligible
    4 (1) --> 4108 (2) not eligible
    Excess = 0
    Queue = 6 5
================
Active node 6 Height = 1 IN = 1 OUT = 1
    6 (1) --> sink (0) pushing 0 from 6 toward sink
    6 (1) --> 4105 (4) not eligible
    Excess = 0
    Queue = 5
================
Active node 5 Height = 1 IN = 1 OUT = 1
    5 (1) --> sink (0) pushing 0 from 5 toward sink
    5 (1) --> 4106 (2) not eligible
    Excess = 0
    Queue =
4

1 回答 1

3

你的问题

预流

没有明确的预流阶段source。它包含在 中，并且在谓词为while时首先执行一次。也许这样做是为了缩短代码P > 0 && v == strue

是的，我认为这样做是为了压缩代码。由于wt[s] = Edge.M*G.V()开始时的分配，源有足够的“虚拟”过剩，以便在算法的第一次迭代中为预流提供燃料。如果你想在你的实现中做同样的技巧excess，当你遇到源节点时，你必须在第一次迭代期间膨胀变量（你即时计算而不是将它存储在像 Sedgewick 这样的数组中）。但是您不必那样做，您的显式预泛洪似乎很好，甚至可能使事情更具可读性。

我还怀疑在这种情况下P > 0 && v == s || h[v] == h[w]+1，隐式运算符优先级(P > 0 && v == s) || h[v] == h[w]+1不是有意的。它的编写方式，P > 0仅针对源案例执行检查。在其他情况下不检查它不会受到伤害，因为执行主体P == 0只会将 0 多余的值推送到其他节点（这不会做任何事情）并且不必要地将这些其他节点添加到队列中，只是为了它们会被立即再次删除. 我已经看到您以相同的方式实现它。没关系，即使稍微浪费了计算时间。大概P > 0 && (v == s || h[v] == h[w]+1)是真的有意为之。

在队列中沉没

根据我的理解和书中的论述，sink 永远不会进入队列。但是，当高度增加时，代码会检查 v != t。这有什么原因吗？

我同意，这很奇怪。接收器进入队列的唯一方法是同时成为源（在只有一个节点且没有边的图中）。但是，在这种情况下，条件v != s已经避免了无限循环，不需要额外的条件。

错误的结果

执行后，产生的流程如下[...] 有了这个流程，流量值为8，但最大值为9

初始高度

你得到错误的结果是因为你的初始高度是错误的。我无法将 Sedgewick 的初始化代码与您的进行比较，因为您的帖子也没有指定。但是我从您的日志文件中了解到，您从 3 的高度开始source。这显然违反了高度函数的条件：source必须以等于图中节点数的高度开始，并将在整个算法期间保持它。

在您的情况下，高度source太接近其下游邻居的高度。这会导致下游邻居很快将一些流量推回源头，然后再努力让它流向下游。只有在没有办法让它流到水槽时，他们才应该这样做。

损坏的图表？

我更担心的是边缘[4104 -> 1]似乎消失了。虽然它在节点 4104 的处理过程中被提及，但它在节点 1 的处理过程中从未出现。所有其他传入边都被提及，无论是关于“不合格”、“推送”还是“减少”。我希望它像其他类型一样显示为三种类型之一。但是话又说回来，我不知道您将这些日志记录语句放在哪里。尽管如此，它还是有点令人担忧，我想我会提到它，以防你在固定高度后仍然遇到问题。

于 2015-06-07T01:51:37.360 回答