Question

这个问题是关于从一堆点中找到共线点。

首先，我不了解slopeMap和无序地图如何？映射不是仅假定具有键和值（map<key, value>）吗？在此特定代码中

unordered_map<pair<int, int>, int,boost:: 
          hash<pair<int, int> > > slopeMap;

据我所知，它具有一对作为键，其后有一个int，应该是值，但是那不就到此为止了吗？

完整代码：

using namespace std; 

// method to find maximum colinear point 
int maxPointOnSameLine(vector< pair<int, int> > points) 
{ 


    int N = points.size(); 
      if (N < 2) 
        return N; 

    int maxPoint = 0; 
    int curMax, overlapPoints, verticalPoints; 

    // here since we are using unordered_map  
    // which is based on hash function  
    //But by default we don't have hash function for pairs 
    //so we'll use hash function defined in Boost library 
    unordered_map<pair<int, int>, int,boost:: 
              hash<pair<int, int> > > slopeMap; 

    // looping for each point 
    for (int i = 0; i < N; i++) 
    { 
        curMax = overlapPoints = verticalPoints = 0; 

        // looping from i + 1 to ignore same pair again 
        for (int j = i + 1; j < N; j++) 
        { 
            // If both point are equal then just 
            // increase overlapPoint count 
            if (points[i] == points[j]) 
                overlapPoints++; 

            // If x co-ordinate is same, then both 
            // point are vertical to each other 
            else if (points[i].first == points[j].first) 
                verticalPoints++; 

            else
            { 
                int yDif = points[j].second - points[i].second; 
                int xDif = points[j].first - points[i].first; 
                int g = __gcd(xDif, yDif); 

                // reducing the difference by their gcd 
                yDif /= g; 
                xDif /= g; 

                // increasing the frequency of current slope 
                // in map 
                slopeMap[make_pair(yDif, xDif)]++; 
                curMax = max(curMax, slopeMap[make_pair(yDif, xDif)]); 
            } 

            curMax = max(curMax, verticalPoints); 
        } 

        // updating global maximum by current point's maximum 
        maxPoint = max(maxPoint, curMax + overlapPoints + 1); 

        // printf("maximum colinear point  
        // which contains current point  
        // are : %d\n", curMax + overlapPoints + 1); 
        slopeMap.clear(); 
    } 

    return maxPoint; 
}

//驱动程序代码

int main() 
{ 
    const int N = 6; 
    int arr[N][2] = {{-1, 1}, {0, 0}, {1, 1}, {2, 2}, 
                    {3, 3}, {3, 4}}; 

    vector< pair<int, int> > points; 
    for (int i = 0; i < N; i++) 
        points.push_back(make_pair(arr[i][0], arr[i][1])); 

    cout << maxPointOnSameLine(points) << endl; 

    return 0; 
}

位置

Input : points[] = {-1, 1}, {0, 0}, {1, 1}, 
                    {2, 2}, {3, 3}, {3, 4} 
Output : 4
Then maximum number of point which lie on same
line are 4, those point are {0, 0}, {1, 1}, {2, 2},
{3, 3}

我还想要一个基于逻辑的建议。我该如何修改此代码，而不是返回定义共线性的最大点数的数字，而是以某种形式的数据结构实际存储共线性的任何点，以便以后使用？

来源： Counting maximum points of the same line

Answer 1

地图不是仅假定具有键和值（地图）吗？

实际上，std::unordered_map还有几个附加的模板参数。第三个参数是哈希。

我该如何修改此代码，而不是返回数字定义共线的最大点数，以实际存储任何我可以使用的某种形式的数据结构中共线的点以后吗？

使用整数坐标比使用浮点数更容易（由于精度问题）。我假设您有一个二维点p的数组。一种方法是，对于每两对点p[i]和p[j]，让您的密钥为对dX，dY 简化为最低形式（其中{{1} }和dX = p[j].x - p[i].x）。那么您的值可以是dY = p[j].y - p[i].y，其中包含匹配的索引std::set<int>和i。

unordered_map是否仅应具有两个参数？

1 个答案: