这个Sedgewick代码是否正确？
我正在解决一个优化问题，除此之外，我必须最大化流网络。我实现了一个基于c ++代码的流量最大化算法，该算法基于以下 java代码，出现在Sedgewick“Java中的算法，第三版，第5部分：图算法”一书中，它最大化了网络流程使用基于顶点的PREFLOW-push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); }
}
}
}
基于该代码，我的实现无法最大化以下网络
执行后，得到的流程如下
通过该流程，流量值为8，但最大值为9，如下图所示
根据我的理解，该算法与本书的解释是一致的。但是，我看到了两件奇怪的事情
- 来源没有明确的预流阶段。它包含在
while中并且在谓词P > 0 && v == s为真时仅执行一次且仅执行一次。也许这是为了缩短代码 - 根据我的理解和本书的话语，接收器永远不会进入队列。但是，当高度增加时，代码会检查v！= t。有什么理由吗？
这是我在C ++中实现此算法的摘录
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的代码之间存在任何逻辑上的不一致吗？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。但我无法理解为什么
我显示了一个详细的执行跟踪，网络无法最大化（跟踪从第一个q.get（）开始。括号中的值是高度的值.IN是到节点的传入流出来的那个。
例如，行
4104 (2) --> 0 (1) pushing 1 from 4104 toward 0
指的是合格弧4104 - > 0。节点4104具有高度2并且节点0具有高度1.表达“推动1”意味着将1个统一的流推向目标节点（0）。行================分隔每个队列提取。队列是FIFO，并且在每次处理结束时打印其状态。
请注意，很多时候推送或减少零流量单位，但目标节点变为活动状态。
这是执行追踪
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 =
1 个答案:
答案 0 :(得分：3)
您的问题
预流
source没有明确的预流阶段。它包含在while中，并且在谓词P > 0 && v == s为true时仅执行一次且仅执行一次。也许这是为了缩短代码
是的，我认为这样做是为了压缩代码。由于在开始时赋值wt[s] = Edge.M*G.V()，源具有足够的“虚拟”过量，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，你将不得不在第一次迭代期间膨胀excess变量（你在飞行中计算而不是将它存储在像Sedgewick这样的数组中），当你遇到源节点。但是你不必这样做，你明确的预装似乎很好，甚至可能使事情更具可读性。
我还怀疑在条件P > 0 && v == s || h[v] == h[w]+1中，隐含的运算符优先级(P > 0 && v == s) || h[v] == h[w]+1不是预期的。编写方式时，仅对源案例执行检查P > 0。在其他情况下不检查它不会有害，因为使用P == 0执行主体只会将0超出其他节点（不执行任何操作），并且不必要地将其他节点添加到队列中，仅用于他们立即被删除。我见过你以同样的方式实现它。即使稍微浪费计算时间，也没关系。可能P > 0 && (v == s || h[v] == h[w]+1)确实是有意的。
汇入队列
根据我的理解和本书的话语，接收器永远不会进入队列。但是，当高度增加时，代码会检查v！= t。有什么理由吗？
我同意，这很奇怪。接收器可以进入队列的唯一方法是同时作为源（在只有一个节点且没有边缘的图形中）。但是，在这种情况下，条件v != s已经避免了无限循环，没有必要的额外条件。
错误的结果
执行后，得到的流程如下[...]使用此流程，流量值为8，但最大值为9
初始高度
您的结果错误，因为您的初始高度是错误的。我无法将Sedgewick的初始化代码与您的初始代码进行比较，因为您的帖子也没有指定。但是我从你的日志文件中了解到，source的高度为3。这显然是针对the conditions for height functions：source必须从图表中的节点数开始，并在整个算法中保留它。
在您的情况下，source的高度太接近其下游邻居的高度。这导致下游邻居很快将一些流量推回到源，然后再努力让它进一步向下游流动。如果没有办法让它流到接收器，它们应该只用 。
破碎图？
我还担心边缘[4104 -> 1]似乎消失了。虽然在节点4104的处理期间提到它，但它在节点1的处理期间从不出现。所有其他进入边缘被提及，无论是关于“不合格”，“推”还是“减少”。我希望它看起来像三种类型中的一种，就像其他类型一样。但话说回来，我不知道你把这些日志记录放在哪里。尽管如此，它还是引起了一些担忧，我想我会提到它，以防你在修好高度后仍然遇到麻烦。
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