通过从两个数组中依次选择元素来获得最大和

给定两个大小为N的数组以及两个整数X和Y，分别表示一个人可以分别从数组A和数组B中选取的最大元素数。
在每个^第i圈，都可以选择A [i]或B [i] 。任务是进行选择，以得到最大可能的和。
注意：保证(X + Y)≥N 。

例子：

Input: A[] = {1, 2, 3, 4, 5}, B[] = {5, 4, 3, 2, 1}, X = 3, Y = 2
Output: 21
i = 0 -> 5 picked
i = 1 -> 4 picked
i = 2 -> 3 picked
i = 3 -> 4 picked
i = 4 -> 5 picked
5 + 4 + 3 + 4 + 5 = 21

Input: A[] = {1, 4, 3, 2, 7, 5, 9, 6}, B[] = {1, 2, 3, 6, 5, 4, 9, 8}, X = 4, Y = 4
Output: 43

方法：使用贪婪方法选择将导致最大总和的元素。以下步骤将有助于解决此问题：

• 根据绝对差对元素对进行排序，即| A [i] – B [i] |以降序排列。
• 比较A [i]和B [i]的值，取较大者将该值加到sum上。
• 如果从A [i]中选择元素，则将X减1，否则将Y减1。
• 最后打印总和。

下面是上述方法的实现：

// Java program to calculate the
// maximum sum obtained by making an
// Optimal selection of elements from two given arrays
  
import java.io.*;
import java.util.*;
import java.lang.*;
  
// User defined Pair class
class Pair {
    int x;
    int y;
  
    // Constructor
    public Pair(int x, int y)
    {
        this.x = x;
        this.y = y;
    }
}
  
// Class to define user defined comparator
class Compare {
  
    // Function to reverse the elements of an array
    static void reverseArray(Pair[] arr, int start, int end)
    {
  
        int tempx, tempy;
        while (start < end) {
            tempx = arr[start].x;
            tempy = arr[start].y;
            arr[start].x = arr[end].x;
            arr[start].y = arr[end].y;
            arr[end].x = tempx;
            arr[end].y = tempy;
            start++;
            end--;
        }
    }
  
    // Function to sort the pair according to the
    // absolute differences
    static Pair[] compare(Pair[] arr, int N)
    {
  
        // Comparator to sort the pair according to 
        // the absolute differences
        Arrays.sort(arr, new Comparator() {
            @Override
            public int compare(Pair p1, Pair p2)
            {
                return (Math.abs(p1.x - p1.y) - Math.abs(p2.x - p2.y));
            }
        });
  
        // To get in descending order
        reverseArray(arr, 0, N - 1);
        return arr;
    }
}
  
// Driver class
class GFG {
  
    // Function to calculate the
    // maximum possible sum obtained by making an
    // optimal selection elements from two given arrays
  
    static int getMaximumSum(int[] A, int[] B, int N,
                                        int X, int Y)
    {
        int num1, num2, sum = 0;
  
        // Making a single pair array having
        // arr[i] element as (Ai, Bi)
        Pair[] arr = new Pair[N];
        for (int i = 0; i < N; i++) {
            arr[i] = new Pair(A[i], B[i]);
        }
  
        // Sorting according to the absolute differences
        // in the decreasing order
        Compare obj = new Compare();
        obj.compare(arr, N);
  
        // Applying Greedy approach to make an optimal 
        // selection
        for (int i = 0; i < N; i++) {
            num1 = arr[i].x;
            num2 = arr[i].y;
  
            // If A[i] > B[i]
            if (num1 > num2) {
  
                // If element from A can be picked
                if (X > 0) {
                    sum += num1;
                    X--;
                }
  
                // Insufficient X
                // Make a pick from B
                else if (Y > 0) {
                    sum += num2;
                    Y--;
                }
            }
  
            // If B[i] > A[i]
            else if (num2 > num1 && Y > 0) {
  
                // If element from B can be picked
                if (Y > 0) {
                    sum += num2;
                    Y--;
                }
  
                // Insufficient Y
                // Make a pick from A
                else {
                    sum += num1;
                    X--;
                }
            }
  
            // If A[i] = B[i]
            // Doesn't make a difference so any value 
            //  can be picked
            else {
                sum += num1;
                if (X > 0) {
                    X--;
                }
                else if (Y > 0)
                    Y--;
            }
        }
        return sum;
    }
  
    // Driver code
    public static void main(String args[])
    {
        int X = 3, Y = 2;
        int[] A = { 1, 2, 3, 4, 5 };
        int[] B = { 5, 4, 3, 2, 1 };
        int N = A.length;
        System.out.println(getMaximumSum(A, B, N, X, Y));
    }
}

输出：

21

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