c++ - 迭代合并std::unordered_map

我有一个节点列表，每个节点都分解为更多节点。例如

Node0 = w01 * Node1 + w02 * Node2 + w03 * Node3的

Node1 = w12 * Node2 + w14 * Node4

因此，我们有Node0 = w01 * w12 * Node2 + w03 * Node3 + w01 * w14 Node4。

我的C++代码用于对给定的一组权重分解执行上述聚合/分解/合并，如下所示。但是，我觉得有很多需要优化的地方。仅举一个例子，我正在遍历topWeights的密钥，并将它们收集在topNodeNames中，这似乎效率很低。

是否有任何STL算法可以帮助我加快速度，并避免不必要的复制？

#include <string>
#include <unordered_map>

template<class T, class U> using umap = std::unordered_map<T, U>;


umap<std::string, double> getWeights(const std::string& nodeName, const umap<std::string, umap<std::string, double>>& weightTrees)
{
    const auto it = weightTrees.find(nodeName);
    if (it == weightTrees.end())
        return umap<std::string, double>();

    umap<std::string, double> topWeights = it->second;
    std::vector<std::string> topNodeNames;

    for (const auto& kv : topWeights)
        topNodeNames.push_back(kv.first);

    for (const std::string& topNodeName : topNodeNames)
    {
        umap<std::string, double> subWeights = getWeights(topNodeName, weightTrees);
        if (subWeights.size() > 0)
        {
            const double topWeight = topWeights[topNodeName];
            topWeights.erase(topNodeName);
            for (const auto& subWeight : subWeights)
            {
                const auto it = topWeights.find(subWeight.first);
                if (it == topWeights.end())
                    topWeights[subWeight.first] = topWeight * subWeight.second;
                else
                    it->second += topWeight * subWeight.second;
            }
        }
    }

    return topWeights;
}


int main()
{
    umap<std::string, umap<std::string, double>> weightTrees = {{ "Node0", {{ "Node1",0.5 },{ "Node2",0.3 },{ "Node3",0.2 }} },
                                                                { "Node1", {{ "Node2",0.1 },{ "Node4",0.9 }} }};

    umap<std::string, double> w = getWeights("Node0", weightTrees); // gives {Node2: 0.35, Node3: 0.20, Node4: 0.45}
}

最佳答案

主要问题是，您将每个节点的递归到每个子节点的，这通常是高度冗余的。避免这种情况的一种方法是在节点名称上引入顺序，其中“较高”的节点仅取决于“较低”的节点，然后以相反的顺序进行计算(对于每个节点，您已经完全知道所有子权重)。但是，我认为没有std算法会为您找到此顺序，因为您不能廉价地临时确定节点依赖性(“节点X是否依赖于节点Y？如果不是直接依赖于节点Y，我们可能必须搜索整个树...”)。

因此，您可以走动态编程路线，并将完全计算出的节点存储在某个位置。甚至更好-您可以在遍历整个树时将其整平为只有叶子的权重。只要您在整个递归中保持展平，实际上在递归形式中这是相当优雅的:

using NodeWeights = std::unordered_map<std::string, double>;
using NonLeaves = std::unordered_map<std::string, NodeWeights>;

// Modifies the tree so that the given root has no non-leaf children.
void flattenTree(std::string root, NonLeaves& toFlatten)
{
    auto rootIt = toFlatten.find(root);
    if (rootIt == toFlatten.end())
        return;

    NodeWeights& rootWeights = rootIt->second;

    NodeWeights leafOnlyWeights;

    for (auto kvp : rootWeights)
    {
        const std::string& childRoot = kvp.first;
        double childWeight = kvp.second;

        std::cout << "Checking child " << childRoot << std::endl;

        // If the graph is indeed acyclic, then the root kvp here is untouched
        // by this call (and thus references to it are not invalidated).
        flattenTree(childRoot, toFlatten);

        auto childIt = toFlatten.find(childRoot);

        // The child is a leaf after flattening: Do not modify anything.
        if (childIt == toFlatten.end())
        {
            leafOnlyWeights[childRoot] = childWeight;
            continue;
        }

        // Child is still not a leaf (but all its children are now leaves):
        // Redistribute its weight among our other child weights.
        const NodeWeights& leafWeights = childIt->second;
        for (auto leafKvp : leafWeights)
            leafOnlyWeights[leafKvp.first] += childWeight * leafKvp.second;
    }

    rootWeights = leafOnlyWeights;
}

int main()
{
    umap<std::string, umap<std::string, double>> weightTrees = {{ "Node0", {{ "Node1",0.5 },{ "Node2",0.3 },{ "Node3",0.2 }} },
                                                                { "Node1", {{ "Node2",0.1 },{ "Node4",0.9 }} }};

    auto flattenedTree = weightTrees;
    flattenTree("Node0", flattenedTree);

    umap<std::string, double> w = flattenedTree["Node0"]; // Should give {Node2: 0.35, Node3: 0.20, Node4: 0.45}

    for (auto kvp : w)
      std::cout << kvp.first << ": " << kvp.second << std::endl;
}

Demo

由于每个节点最多只能展平一次，因此您将无法运行原始算法具有的指数运行时间。