是不是unordered_map只应该有两个参数？

Question

这个问题是关于从一堆点中找到共线点。

首先，我不明白slopeMap和无序地图怎么样？不是地图只假设有一个键和一个值（map<key, value>）？在这个特定的代码中

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

根据我的理解，它有一对作为键和一个后面的int，它应该是值，但那么它并不完全在那里结束？

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; 
} 

哪里

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。我怎么能修改这个代码，以便不是返回一个定义最大点数colinear的数字，而是实际存储我以后可能使用的某种形式的数据结构的共线点？

资料来源：Counting maximum points of the same line

Answer 1

不是地图只假设有一个键和一个值（地图）？

实际上，std::unordered_map有几个额外的模板参数。第三个参数是哈希。

我怎么能修改这个代码，以便不是返回一个定义最大点数colinear的数字，而是实际存储我以后可能使用的某种形式的数据结构的共线点？

使用整数坐标比使用浮点更容易（由于精度问题）。我假设你有一个2D点p阵列。一种方法是，对于每两对点p[i]和p[j]，让你的密钥成对dX，dY减少到最低形式（其中dX = p[j].x - p[i].x和dY = p[j].y - p[i].y）。然后你的价值可能是一个std::set<int>，其中包含匹配的指数i和j。