从给定的一组线中找到给定 X 的 Y 的最大值

给定由分别由斜率(m)和截距(c)组成的二维数组 arr表示的一组线，以及Q查询，使得每个查询都包含一个值x 。任务是从所有给定的一组行中为每个x值找到 y 的最大值。

The given lines are represented by the equation y = m*x + c.

例子：

Input: arr[][2] ={ {1, 1}, {0, 0}, {-3, 3} }, Q = {-2, 2, 1}
Output: 9, 3, 2
For query x = -2, y values from the equations are -1, 0, 9. So the maximum value is 9
Similarly, for x = 2, y values are 3, 0, -3. So the maximum value is 3
And for x = 1, values of y = 2, 0, 0. So the maximum value is 2.

Input: arr[][] ={ {5, 6}, {3, 2}, {7, 3} }, Q = { 1, 2, 30 }
Output: 10, 17, 213

朴素的方法：朴素的方法是在每一行中替换 x的值并计算所有行中的最大值。对于每个查询，将花费O(N)时间，因此解决方案的复杂性变为O(Q * N) ，其中N是行数， Q是查询数。

有效的方法：想法是使用凸包技巧：

• 从给定的行集合中，可以找到并删除没有意义的行(对于 x 的任何值，它们永远不会给出最大值y )，从而减少集合。
• 现在，如果可以找到每行给出最大值的范围 (l, r) ，则可以使用二分搜索来回答每个查询。
• 因此，创建了一个按斜率递减顺序排列的直线向量，并按斜率递减顺序插入这些直线。

下面是上述方法的实现：

// C++ implementation of
// the above approach
  
#include 
using namespace std;
  
struct Line {
    int m, c;
  
public:
    // Sort the line in decreasing
    // order of their slopes
    bool operator<(Line l)
    {
  
        // If slopes aren't equal
        if (m != l.m)
            return m > l.m;
  
        // If the slopes are equal
        else
            return c > l.c;
    }
  
    // Checks if line L3 or L1 is better than L2
    // Intersection of Line 1 and
    // Line 2 has x-coordinate (b1-b2)/(m2-m1)
    // Similarly for Line 1 and
    // Line 3 has x-coordinate (b1-b3)/(m3-m1)
    // Cross multiplication will
    // give the below result
    bool check(Line L1, Line L2, Line L3)
    {
        return (L3.c - L1.c) * (L1.m - L2.m)
               < (L2.c - L1.c) * (L1.m - L3.m);
    }
};
  
struct Convex_HULL_Trick {
  
    // To store the lines
    vector l;
  
    // Add the line to the set of lines
    void add(Line newLine)
    {
  
        int n = l.size();
  
        // To check if after adding the new line
        // whether old lines are
        // losing significance or not
        while (n >= 2
               && newLine.check(l[n - 2],
                                l[n - 1],
                                newLine)) {
            n--;
        }
  
        l.resize(n);
  
        // Add the present line
        l.push_back(newLine);
    }
  
    // Function to return the y coordinate
    // of the specified line
    // for the given coordinate
    int value(int in, int x)
    {
        return l[in].m * x + l[in].c;
    }
  
    // Function to Return the maximum value
    // of y for the given x coordinate
    int maxQuery(int x)
    {
        // if there is no lines
        if (l.empty())
            return INT_MAX;
  
        int low = 0,
            high = (int)l.size() - 2;
  
        // Binary search
        while (low <= high) {
            int mid = (low + high) / 2;
  
            if (value(mid, x)
                < value(mid + 1, x))
                low = mid + 1;
            else
                high = mid - 1;
        }
  
        return value(low, x);
    }
};
  
// Driver code
int main()
{
    Line lines[] = { { 1, 1 },
                     { 0, 0 },
                     { -3, 3 } };
    int Q[] = { -2, 2, 1 };
    int n = 3, q = 3;
    Convex_HULL_Trick cht;
  
    // Sort the lines
    sort(lines, lines + n);
  
    // Add the lines
    for (int i = 0; i < n; i++)
        cht.add(lines[i]);
  
    // For each query in Q
    for (int i = 0; i < q; i++) {
        int x = Q[i];
        cout << cht.maxQuery(x) << endl;
    }
  
    return 0;
}

输出：

9
3
2

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live