当第一个键发生变化时，获取连续映射的大小

Question

我有一张map<pair<int, int>, int> counts对的地图。我正在迭代这个并且我正在计算一些浮点数，如果地图的值大于2.现在我想将浮点数保存在一个向量中，该向量与我的第一个元素中的相似值的数量一样大。我的一对。我的地图中值的示例：

1 1 4 1 2 5 1 3 5 1 7 5 1 29 2 ..现在第一个键改变，所以大小将是4（不是5，因为一个值<3） 2 10 1 2 20 4 ...

我试着做一个最小的例子，这解释得更好一点：

int main(int argc, char** argv)
{
    // Create some values - both vectors have the same size
    vector<pair<int, int>> pairs;
    pairs.push_back(make_pair(1, 1));
    pairs.push_back(make_pair(1, 2));
    pairs.push_back(make_pair(1, 4));
    pairs.push_back(make_pair(2, 7));
    pairs.push_back(make_pair(2, 4));
    pairs.push_back(make_pair(3, 5));
    pairs.push_back(make_pair(3, 7));
    pairs.push_back(make_pair(3, 8));

    vector<float> value;
    value.push_back(4);
    value.push_back(5);
    value.push_back(8);
    value.push_back(2);
    value.push_back(5);
    value.push_back(6);
    value.push_back(7);
    value.push_back(8);

    map<pair<int, int>, int> counts;
    vector<vector<vector<float>>> descriptors_1, descriptors_2;
    vector<float> oa_simil;
    float Overall_Similarity;

    for (size_t i = 0; i < pairs.size(); i++) {
        counts.insert(make_pair(make_pair(pairs[i].first, pairs[i].second), value[i]));
    }

    for (const auto& p : counts) {
        const auto& p1 = p.first.first;
        const auto& p2 = p.first.second;
        int count = p.second;
        if (p.second >= 3) {
            float S = 0;
            // Two for-loops that calculate a new S for every p1, p2 combination >= 3
            //S = ls_1;
            // 5 Linesegments in image 1 are compared to 5 Linesegments in image 2, if p.second >= 3
            for (int ls_1 = 0; ls_1 < 5; ls_1++) {  
                for (int ls_2 = 0; ls_2 < 5; ls_2++) {
                    pair<int, int> index_1, index_2;
                    index_1 = make_pair(p1, ls_1);
                    index_2 = make_pair(p2, ls_2);

                    // Calculate the Similarity of different pairs of line segments in a complex function
                    calculateSimilarity(Overall_Similarity, descriptors_1, descriptors_2, index_1, index_2);
                    oa_simil.push_back(Overall_Similarity);
                }
                // Get the maximum similarity of every single line segment in one image
                float max_M = *max_element(oa_simil.begin(), oa_simil.end());
                oa_simil.clear();

                // Sum up the maxima -> S contains the maximum for the combination of two Linesegments p1, p2 (if p >=3)
                S += max_M;
            }

            // Now I want to get the maximum value S for every single p1
            // In the example I want the maximum S out of e.g. the 3 pairs (1,1), (1,2), (1,4).
            // Trying to push the S in a vector<float> when p1 changes.

        }

    }

}

我已经尝试将所有的键和S的值放入一个新的地图中，但是当我已经在迭代我的键时，这似乎浪费了计算时间。

Answer 1

我想我解决了。我为我的对和S值创建了一个地图，并反转了结果，以便查找最大值。然后我将我的p1与前一次迭代的p1进行比较，以测试该值是否发生了变化。

template<typename A, typename B>
std::pair<B, A> flip_pair(const std::pair<A, B> &p)
{
    return std::pair<B, A>(p.second, p.first);
}

template<typename A, typename B>
std::multimap<B, A> flip_map(const std::map<A, B> &src)
{
    std::multimap<B, A> dst;
    std::transform(src.begin(), src.end(), std::inserter(dst, dst.begin()),
        flip_pair<A, B>);
    return dst;
}

int main(int argc, char** argv) 
{

map<pair<int, int>, float> similarities;
multimap<float, pair<int, int>> inversed_map;

...             

               // Get the maximum similarity of every single line segment in one image
                float max_M = *max_element(oa_simil.begin(), oa_simil.end());
                oa_simil.clear();
                // Sum up the maxima -> S contains the maximum for the combination of two Linesegments p1, p2 (if p >=3)
                S += max_M;
            }


            similarities.insert(make_pair(make_pair(p1, p2), S));
            inversed_map = flip_map(similarities);

            if ((p1 > p1_old) && ((int)inversed_map.size() > 0)) {
                // Sort the vector and keep track of the indexes of the pair.
                auto it = inversed_map.rbegin();
                cout << it->second.first << " " << it->second.second << " " << it->first << endl;

                similarities.clear();
                inversed_map.clear();

            }
            p1_old = p1;


        }

    }

}