二维平面上任意一对坐标的给定表达式的最大值

给定一个大小为N的排序二维数组arr[][2]使得(arr[i][0], arr[i][1])表示笛卡尔平面中^第i^个点的坐标和一个整数K ，则任务是找到表达式(|arr[i][0] – arr[j][0]| + arr[i][1] + arr[j][1])使得|arr[i][0] – arr[j][0]|对于任何可能的坐标对(i, j)， ≤ K。

例子：

Input: arr[][] = {{1, 3}, {2, 0}, {5, 10}, {6, -10}}, K = 1
Output: 4
Explanation:
Choose pairs (0, 1). Now the value of the expression is given by:
value = (abs(1 – 2) + 3 + 0) = 4, which is maximum and abs(1 – 2) = 1(≤ K).
Therefore, print 4.

Input: arr[][] = {{0, 0}, {3, 0}, {9, 2}}, K = 3
Output: 3

方法：可以使用基于以下观察的优先队列使用贪心算法解决给定的问题：

• 将所有i > j的表达式重新排列为(arr[i][0] – arr[i][1] + arr[j][0] + arr[j][1]) 。
• 现在，保持{arr[i] _x – arr[i] _y , arr[i] _x }对的排序顺序，可以计算索引j处每个数组元素的给定表达式的值。

请按照以下步骤解决问题：

• 初始化成对的priority_queue 说PQ ，它存储了一个点的坐标轴和该点的X坐标的差值对。
• 初始化一个变量res为 INT_MIN 以存储最大值。
• 遍历数组arr[][]并考虑{X, Y}是当前点执行以下操作：
  • 在PQ不为空且(X – PQ.top()[1])大于K 时进行迭代，并从 priority_queue PQ 中删除顶部元素。
  • 如果PQ不为空，则将res的值更新为res和PQ.top()[0] + X + Y)的最大值。
  • 将{Y – X, X}对推入 priority_queue PQ 。
• 完成上述步骤后，打印res的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the maximum value
// of the given expression possible
// for any pair of co-ordinates
void findMaxValueOfEquation(
    vector >& arr, int K)
{
    // Stores the differences between pairs
    priority_queue > pq;
 
    // Stores the maximum value
    int res = INT_MIN;
 
    // Traverse the array arr[][]
    for (auto point : arr) {
 
        // While pq is not empty and
        // difference between point[0]
        // and pq.top()[1] > K
        while (!pq.empty()
               && point[0] - pq.top()[1]
                      > K) {
 
            // Removes the top element
            pq.pop();
        }
 
        // If pq is not empty
        if (!pq.empty()) {
 
            // Update the value res
            res = max(res,
                      pq.top()[0] + point[0] + point[1]);
        }
 
        // Push pair {point[1] - point[0],
        // point[0]} in pq
        pq.push({ point[1] - point[0],
                  point[0] });
    }
 
    // Print the result
    cout << res;
}
 
// Driver Code
int main()
{
    vector > arr
        = { { 1, 3 }, { 2, 0 },
            { 5, 10 }, { 6, -10 } };
    int K = 1;
    findMaxValueOfEquation(arr, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG
{
 
    // Function to find the maximum value
    // of the given expression possible
    // for any pair of co-ordinates
    static void findMaxValueOfEquation(int arr[][], int K)
    {
       
        // Stores the differences between pairs
        PriorityQueue pq
            = new PriorityQueue<>((a, b) -> {
                  if (a[0] != b[0])
                      return b[0] - a[0];
                  return b[1] - a[1];
              });
 
        // Stores the maximum value
        int res = Integer.MIN_VALUE;
 
        // Traverse the array arr[][]
        for (int point[] : arr) {
 
            // While pq is not empty and
            // difference between point[0]
            // and pq.top()[1] > K
            while (!pq.isEmpty()
                   && point[0] - pq.peek()[1] > K) {
 
                // Removes the top element
                pq.poll();
            }
 
            // If pq is not empty
            if (!pq.isEmpty()) {
 
                // Update the value res
                res = Math.max(res, pq.peek()[0] + point[0]
                                        + point[1]);
            }
 
            // Push pair {point[1] - point[0],
            // point[0]} in pq
            pq.add(new int[] { point[1] - point[0],
                               point[0] });
        }
 
        // Print the result
        System.out.println(res);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int[][] arr
            = { { 1, 3 }, { 2, 0 }, { 5, 10 }, { 6, -10 } };
        int K = 1;
        findMaxValueOfEquation(arr, K);
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to find the maximum value
# of the given expression possible
# for any pair of co-ordinates
def findMaxValueOfEquation(arr, K):
   
    # Stores the differences between pairs
    pq = []
 
    # Stores the maximum value
    res = -10**8
 
    # Traverse the array arr[][]
    for point in arr:
 
        # While pq is not empty and
        # difference between point[0]
        # and pq.top()[1] > K
        while (len(pq)>0 and point[0] - pq[-1][1] > K):
           
            # Removes the top element
            del pq[-1]
 
        # If pq is not empty
        if (len(pq) > 0):
            # Update the value res
            res = max(res, pq[-1][0] + point[0] + point[1])
 
        # Push pair {point[1] - point[0],
        # point[0]} in pq
        pq.append([point[1] - point[0], point[0]])
        pq = sorted(pq)
 
    # Prthe result
    print (res)
 
# Driver Code
if __name__ == '__main__':
    arr = [ [ 1, 3 ], [ 2, 0 ], [ 5, 10 ], [ 6, -10 ] ]
    K = 1
    findMaxValueOfEquation(arr, K)
 
# This code is contributed by mohit kumar 29.

输出：

4

时间复杂度： O(N * log N)
辅助空间： O(N)

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