当第一个密钥更改时，将对持续的映射保持持续

我有一个对 map<pair<int, int>, int> counts的映射。我正在迭代此处，并且我正在计算一些浮点数，如果地图的值大于2。我的一对。我地图中值的示例：

1 1 4
1 2 5
1 3 5
1 7 5
1 29 2 ..现在的第一个键更改，因此大小为4(不是5，因为一个值是＆lt; 3(

2 10 1
2 20 4 ...

我试图以最小的示例来解释它：

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.
        }
    }
}

我已经试图将我的所有键和S的价值纳入新地图，但是当我已经通过键迭代时，这似乎是浪费计算时间。