分配给数字连续和不同数组元素的子序列的最大分数

给定两个大小为N 的数组arr[]和brr[] ，使得数组brr[]由与数组arr[] 的相应元素相关联的分数组成。任务是找到数组arr[]的数字连续和不同元素的子序列的分配分数的最大可能总和。
例子：

Input: arr[] = {1, 2, 3, 3, 3, 1}, brr[] = {-1, 2, 10, 20, -10, -9} 
Output: 22 
Explanation: 
Distinct values from the array = {1, 2, 3} 
Maximum value assigned to each element = {1: -1, 2: 2, 3: 20} 
Select the elements at index 2 and 4 in arr[] which are {2, 3}. 
Maximum score = 2 + 20 = 22.

Input: arr[] = {1, 2, 3, 2, 3, 1}, brr[] = {-1, 2, 10, 20, -10, -9} 
Output: 32 
Explanation: Selected subsequence is {arr[1], arr[2], arr[3]} = {2, 3, 2}

朴素方法：解决问题的最简单方法是使用递归。请按照以下步骤解决问题：

1. 生成给定数组arr[] 的所有可能子集，使得该子集具有唯一且连续的元素。
2. 在上述步骤中生成子集时，每个元素是否添加到子序列都有两种可能性。因此，请按照以下步骤操作：
   • 如果当前元素与先前选择的元素相差 1，则将该元素添加到子序列中。
   • 否则，继续下一个元素。
3. 通过考虑以上两种可能性来更新maximum_score 。
4. 打印完整遍历数组后得到的maximum_score的最终值。

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 score
// with unique element in the subset
int maximumSum(int a[], int b[], int n,
               int index, int lastpicked)
{
     
    // Base Case
    if (index == n)
        return 0;
  
    int option1 = 0, option2 = 0;
  
    // Check if the previously picked
    // element differs by 1 from the
    // current element
    if (lastpicked == -1 ||
      a[lastpicked] != a[index])
  
        // Calculate score by including
        // the current element
        option1 = b[index] + maximumSum(
                             a, b, n,
                             index + 1,
                             index);
  
    // Calculate score by excluding
    // the current element
    option2 = maximumSum(a, b, n,
                         index + 1,
                         lastpicked);
  
    // Return maximum of the
    // two possibilities
    return max(option1, option2);
}
  
// Driver code
int main()
{
     
    // Given arrays
    int arr[] = { 1, 2, 3, 3, 3, 1 };
    int brr[] = { -1, 2, 10, 20, -10, -9 };
     
    int N = sizeof(arr) / sizeof(arr[0]);
     
    // Function call
    cout << (maximumSum(arr, brr, N, 0, -1));
}
 
// This code is contributed by rutvik_56

Java

// Java program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to find the maximum score
    // with unique element in the subset
    public static int maximumSum(
        int[] a, int[] b, int n, int index,
        int lastpicked)
    {
        // Base Case
        if (index == n)
            return 0;
 
        int option1 = 0, option2 = 0;
 
        // Check if the previously picked
        // element differs by 1 from the
        // current element
        if (lastpicked == -1
            || a[lastpicked] != a[index])
 
            // Calculate score by including
            // the current element
            option1
                = b[index]
                  + maximumSum(a, b, n,
                               index + 1,
                               index);
 
        // Calculate score by excluding
        // the current element
        option2 = maximumSum(a, b, n,
                             index + 1,
                             lastpicked);
 
        // Return maximum of the
        // two possibilities
        return Math.max(option1, option2);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given arrays
        int arr[] = { 1, 2, 3, 3, 3, 1 };
        int brr[] = { -1, 2, 10, 20, -10, -9 };
 
        int N = arr.length;
 
        // Function Call
        System.out.println(
            maximumSum(arr, brr,
                       N, 0, -1));
    }
}

Python3

# Python3 program for the above approach
 
# Function to find the maximum score
# with unique element in the subset
def maximumSum(a, b, n, index, lastpicked):
     
    # Base Case
    if (index == n):
        return 0
 
    option1 = 0
    option2 = 0
 
    # Check if the previously picked
    # element differs by 1 from the
    # current element
    if (lastpicked == -1 or
      a[lastpicked] != a[index]):
 
        # Calculate score by including
        # the current element
        option1 = b[index] + maximumSum(a, b, n,
                                        index + 1,
                                        index)
 
    # Calculate score by excluding
    # the current element
    option2 = maximumSum(a, b, n, index + 1,
                         lastpicked)
 
# Return maximum of the
# two possibilities
    return max(option1, option2)
 
# Driver Code
if __name__ == '__main__':
     
    # Given arrays
    arr = [ 1, 2, 3, 3, 3, 1 ]
    brr = [ -1, 2, 10, 20, -10, -9 ]
 
    N = len(arr)
 
    # Function call
    print(maximumSum(arr, brr, N, 0, -1))
 
# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to find the maximum score
// with unique element in the subset
public static int maximumSum(int[] a, int[] b,
                             int n, int index,
                             int lastpicked)
{
  // Base Case
  if (index == n)
    return 0;
 
  int option1 = 0, option2 = 0;
 
  // Check if the previously picked
  // element differs by 1 from the
  // current element
  if (lastpicked == -1 ||
      a[lastpicked] != a[index])
 
    // Calculate score by including
    // the current element
    option1 = b[index] + maximumSum(a, b, n,
                                    index + 1,
                                    index);
 
  // Calculate score by excluding
  // the current element
  option2 = maximumSum(a, b, n,
                       index + 1,
                       lastpicked);
 
  // Return maximum of the
  // two possibilities
  return Math.Max(option1, option2);
}   
 
// Driver Code
public static void Main(String[] args)
{
  // Given arrays
  int []arr = {1, 2, 3, 3, 3, 1};
  int []brr = {-1, 2, 10, 20, -10, -9};
 
  int N = arr.Length;
 
  // Function Call
  Console.WriteLine(maximumSum(arr, brr,
                               N, 0, -1));
}
}
 
// This code is contributed by shikhasingrajput

Javascript

 
// JavaScript program to implement
// the above approach
 
    // Function to find the maximum score
    // with unique element in the subset
    function maximumSum(
        a, b, n, index,
        lastpicked)
    {
        // Base Case
        if (index == n)
            return 0;
  
        let option1 = 0, option2 = 0;
  
        // Check if the previously picked
        // element differs by 1 from the
        // current element
        if (lastpicked == -1
            || a[lastpicked] != a[index])
  
            // Calculate score by including
            // the current element
            option1
                = b[index]
                  + maximumSum(a, b, n,
                               index + 1,
                               index);
  
        // Calculate score by excluding
        // the current element
        option2 = maximumSum(a, b, n,
                             index + 1,
                             lastpicked);
  
        // Return maximum of the
        // two possibilities
        return Math.max(option1, option2);
    }
  
// Driver code
 
        // Given arrays
        let arr = [ 1, 2, 3, 3, 3, 1 ];
        let brr = [ -1, 2, 10, 20, -10, -9 ];
  
        let N = arr.length;
  
        // Function Call
        document.write(
            maximumSum(arr, brr,
                       N, 0, -1));
 
// This code is contributed by target_2.

输出：

22

时间复杂度： O(2 ^N )
辅助空间： O(1)

有效的方法：由于问题具有重叠子问题，因此可以使用动态规划来优化上述方法。以下是步骤：

1. 将index初始化为0并将lastPicked初始化为-1 。
2. 初始化一个二维数组，比如dp[][]来存储子问题的结果。
3. dp[][] 的状态将是当前索引和最后选择的整数。
4. 计算两个可能选项的分数：
   • 如果最后选择的整数与当前整数不同，则选择当前元素。
   • 跳过当前元素移动到下一个元素。
5. 将当前状态存储为上面两个状态计算值的最大值。
6. 在所有递归调用结束后，打印dp[index][lastPicked + 1] 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
// score possible
int maximumSum(int a[], int b[], int n,
               int index, int lastpicked,
               vector> dp)
{
     
    // Base Case
    if (index == n)
        return 0;
 
    // If previously occurred subproblem
    // occurred
    if (dp[index][lastpicked + 1] != -1)
        return dp[index][lastpicked + 1];
 
    int option1 = 0, option2 = 0;
 
    // Check if lastpicked element differs
    // by 1 from the current element
    if (lastpicked == -1 ||
      a[lastpicked] != a[index])
    {
         
        // Calculate score by including
        // the current element
        option1 = b[index] + maximumSum(a, b, n,
                                        index + 1,
                                        index, dp);
    }
 
    // Calculate score by excluding
    // the current element
    option2 = maximumSum(a, b, n,
                         index + 1,
                         lastpicked, dp);
 
    // Return maximum score from
    // the two possibilities
    return dp[index][lastpicked + 1] = max(option1,
                                           option2);
}
 
// Function to print maximum score
void maximumPoints(int arr[], int brr[], int n)
{
    int index = 0, lastPicked = -1;
 
    // DP array to store results
    // Initialise dp with -1
    vector> dp(n + 5, vector(n + 5, -1));
 
    // Function call
    cout << maximumSum(arr, brr, n, index,
                       lastPicked, dp)
         << endl;
}
  
// Driver code   
int main()
{
     
    // Given arrays
    int arr[] = { 1, 2, 3, 3, 3, 1 };
    int brr[] = { -1, 2, 10, 20, -10, -9 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    maximumPoints(arr, brr, N);
 
   return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function to find the maximum
    // score possible
    public static int maximumSum(
        int[] a, int[] b, int n, int index,
        int lastpicked, int[][] dp)
    {
        // Base Case
        if (index == n)
            return 0;
 
        // If previously occurred subproblem
        // occurred
        if (dp[index][lastpicked + 1] != -1)
            return dp[index][lastpicked + 1];
 
        int option1 = 0, option2 = 0;
 
        // Check if lastpicked element differs
        // by 1 from the current element
        if (lastpicked == -1
            || a[lastpicked] != a[index]) {
 
            // Calculate score by including
            // the current element
            option1 = b[index]
                      + maximumSum(a, b, n,
                                   index + 1,
                                   index, dp);
        }
 
        // Calculate score by excluding
        // the current element
        option2 = maximumSum(a, b, n,
                             index + 1,
                             lastpicked, dp);
 
        // Return maximum score from
        // the two possibilities
        return dp[index][lastpicked + 1]
            = Math.max(option1, option2);
    }
 
    // Function to print maximum score
    public static void maximumPoints(
        int arr[], int brr[], int n)
    {
        int index = 0, lastPicked = -1;
 
        // DP array to store results
        int dp[][] = new int[n + 5][n + 5];
 
        // Initialise dp with -1
        for (int i[] : dp)
            Arrays.fill(i, -1);
 
        // Function call
        System.out.println(
            maximumSum(arr, brr, n,
                       index, lastPicked, dp));
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given arrays
        int arr[] = { 1, 2, 3, 3, 3, 1 };
        int brr[] = { -1, 2, 10, 20, -10, -9 };
 
        int N = arr.length;
 
        // Function Call
        maximumPoints(arr, brr, N);
    }
}

蟒蛇3

# Python3 program for
# the above approach
 
# Function to find the
# maximum score possible
def maximumSum(a, b, n, index,
               lastpicked, dp):
 
    # Base Case
    if (index == n):
        return 0
 
    # If previously occurred
    # subproblem occurred
    if (dp[index][lastpicked + 1] != -1):
        return dp[index][lastpicked + 1]
 
    option1, option2 = 0, 0
 
    # Check if lastpicked element differs
    # by 1 from the current element
    if (lastpicked == -1 or
        a[lastpicked] != a[index]):
 
        # Calculate score by including
        # the current element
        option1 = (b[index] +
                   maximumSum(a, b, n,
                              index + 1,
                              index, dp))
    
    # Calculate score by excluding
    # the current element
    option2 = maximumSum(a, b, n,
                         index + 1,
                         lastpicked, dp)
 
    # Return maximum score from
    # the two possibilities
    dp[index][lastpicked + 1] = max(option1,
                                    option2)
    return dp[index][lastpicked + 1]
 
# Function to print maximum score
def maximumPoints(arr, brr, n):
 
    index = 0
    lastPicked = -1
 
    # DP array to store results
    dp =[[ -1 for x in range (n + 5)]
              for y in range (n + 5)]
 
    # Function call
    print (maximumSum(arr, brr,
                      n, index,
                      lastPicked, dp))
 
# Driver Code
if __name__ == "__main__":
   
    # Given arrays
    arr = [1, 2, 3, 3, 3, 1]
    brr = [-1, 2, 10, 20, -10, -9]
 
    N = len(arr)
 
    # Function Call
    maximumPoints(arr, brr, N)
 
# This code is contributed by Chitranayal

C＃

// C# program for
// the above approach
using System;
class GFG{
 
// Function to find the maximum
// score possible
public static int maximumSum(int[] a, int[] b,
                             int n, int index,
                             int lastpicked,
                             int[,] dp)
{
  // Base Case
  if (index == n)
    return 0;
 
  // If previously occurred
  // subproblem occurred
  if (dp[index, lastpicked + 1] != -1)
    return dp[index, lastpicked + 1];
 
  int option1 = 0, option2 = 0;
 
  // Check if lastpicked element differs
  // by 1 from the current element
  if (lastpicked == -1 ||
      a[lastpicked] != a[index])
  {
    // Calculate score by including
    // the current element
    option1 = b[index] + maximumSum(a, b, n,
                                    index + 1,
                                    index, dp);
  }
 
  // Calculate score by excluding
  // the current element
  option2 = maximumSum(a, b, n,
                       index + 1,
                       lastpicked, dp);
 
  // Return maximum score from
  // the two possibilities
  return dp[index, lastpicked + 1] =
         Math.Max(option1, option2);
}
 
// Function to print maximum score
public static void maximumPoints(int []arr,
                                 int []brr,
                                 int n)
{
  int index = 0, lastPicked = -1;
 
  // DP array to store results
  int [,]dp = new int[n + 5, n + 5];
 
  // Initialise dp with -1
  for(int i = 0; i < n + 5; i++)
  {
    for (int j = 0; j < n + 5; j++)
    {
      dp[i, j] = -1;
    }
  }
  // Function call
  Console.WriteLine(maximumSum(arr, brr, n,
                               index, lastPicked,
                               dp));
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given arrays
  int []arr = {1, 2, 3, 3, 3, 1};
  int []brr = {-1, 2, 10, 20, -10, -9};
 
  int N = arr.Length;
 
  // Function Call
  maximumPoints(arr, brr, N);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program for the above approach
 
    // Function to find the maximum
    // score possible
    function maximumSum(
        a, b, n, index,
        lastpicked, dp)
    {
        // Base Case
        if (index == n)
            return 0;
  
        // If previously occurred subproblem
        // occurred
        if (dp[index][lastpicked + 1] != -1)
            return dp[index][lastpicked + 1];
  
        let option1 = 0, option2 = 0;
  
        // Check if lastpicked element differs
        // by 1 from the current element
        if (lastpicked == -1
            || a[lastpicked] != a[index]) {
  
            // Calculate score by including
            // the current element
            option1 = b[index]
                      + maximumSum(a, b, n,
                                   index + 1,
                                   index, dp);
        }
  
        // Calculate score by excluding
        // the current element
        option2 = maximumSum(a, b, n,
                             index + 1,
                             lastpicked, dp);
  
        // Return maximum score from
        // the two possibilities
        return dp[index][lastpicked + 1]
            = Math.max(option1, option2);
    }
  
    // Function to prlet maximum score
    function maximumPolets(
        arr, brr, n)
    {
        let index = 0, lastPicked = -1;
  
        // DP array to store results
        let dp = new Array(n + 5);
        // Loop to create 2D array using 1D array
        for (var i = 0; i < dp.length; i++) {
            dp[i] = new Array(2);
        }
  
        // Initialise dp with -1
        for (var i = 0; i < dp.length; i++) {
            for (var j = 0; j < dp.length; j++) {
            dp[i][j] = -1;
        }
        }
  
        // Function call
        document.write(
            maximumSum(arr, brr, n,
                       index, lastPicked, dp));
    }
  
// Driver Code
 
   // Given arrays
        let arr = [ 1, 2, 3, 3, 3, 1 ];
        let brr = [ -1, 2, 10, 20, -10, -9 ];
  
        let N = arr.length;
  
        // Function Call
        maximumPolets(arr, brr, N);
           

输出：

22

时间复杂度： O(N ² )
辅助空间： O(N ² )

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