给定两个数组的最小总和

给定N个正整数的两个数组A []和B []和代价C。我们可以从给定数组的每个索引中选择任意一个元素，即，对于任何索引i，我们只能选择元素A [i]或B [i] 。任务是找到从给定的两个数组中选择N个元素的最小总和，如果我们在下一次迭代中选择从A []到B []或反之亦然的任何元素，那么成本C也将添加到和。
注意：按索引的升序选择元素，即0≤i 。

例子：

Input: N = 9, A[] = {7, 6, 18, 6, 16, 18, 1, 17, 17}, B[] = {6, 9, 3, 10, 9, 1, 10, 1, 5}, C = 2 
Output: 49 
Explanation: 
On taking the 1st element from array A, sum = 7 
On taking the 2nd element from array A, sum = 7 + 6 = 13 
On taking the 3rd element from array B, as we are entering from array A to array B, sum = 13 + 3 + 2 = 18 
On taking the 4th element from array A, as we are entering from array B to array A, sum = 18 + 6 + 2 = 26 
On taking the 5th element from array B, as we are entering from array A to array B, sum = 26 + 9 + 2 = 37 
On taking the 6th element form array B, sum = 37 + 1 = 38 
On taking the 7th element from array A, as we are entering from array B to array A, sum = 38 + 1 + 2 = 41 
On taking the 8th element form array B, as we are entering from array A to array B, sum = 41 + 1 + 2 = 44 
On taking the 9th element from array B, sum = 44 + 5 = 49.

Input: N = 9, A = {3, 2, 3, 1, 3, 3, 1, 4, 1}, B = {1, 2, 3, 4, 4, 1, 2, 1, 3}, C = 1 
Output: 18 
Explanation: 
On taking the 1st element from array B, sum = 1 
On taking the 2nd element from array A, sum = 1 + 2 = 3 
On taking the 3rd element from array A, sum = 3 + 3 = 6 
On taking the 4th element from array A, sum = 6 + 1 = 7 
On taking the 5th element from array A, sum = 7 + 3 = 10 
On taking the 6th element form array B, as we are entering from array A to array B, sum = 10 + 1 + 1 = 12 
On taking the 7th element from array A, as we are entering from array B to array A, sum = 12 + 1 + 1 = 14 
On taking the 8th element form array B, as we are entering from array A to array B, sum = 14 + 1 + 1 = 16 
On taking the 9th element from array A, as we are entering from array B to array A, sum = 16 + 1 + 1 = 18. 

方法：我们将使用动态编程来解决此问题。步骤如下：

1. 创建N行两列的2D数组dp [] []并将dp的所有元素初始化为infinity 。
2. 在两个数组中添加元素可能有4种情况：
   • 当先前添加的元素来自数组a []时，从数组a []添加元素。
   • 当先前添加的元素来自数组b []时，从数组a []添加元素。在这种情况下，将整数C与结果相加是有代价的。
   • 当先前添加的元素来自数组b []时，从数组b []添加元素。
   • 当先前添加的元素来自数组a []时，从数组b []添加元素。在这种情况下，将整数C与结果相加是有代价的。
3. 每次使用上述四个条件的最小值更新dp数组。
4. dp [n-1] [0]和dp [n-1] [1]的最小值是选择N个元素的总和。

下面是上述方法的实现：

C++

// C++ program for the above approach 
#include  
using namespace std; 
  
// Function that prints minimum sum 
// after selecting N elements 
void minimumSum(int a[], int b[], 
                int c, int n) 
{ 
  
    // Initialise the dp array 
    vector > dp(n, 
                            vector(2, 
                                        1e6)); 
  
    // Base Case 
    dp[0][0] = a[0]; 
    dp[0][1] = b[0]; 
  
    for (int i = 1; i < n; i++) { 
  
        // Adding the element of array a if 
        // previous element is also from array a 
        dp[i][0] = min(dp[i][0], 
                    dp[i - 1][0] + a[i]); 
  
        // Adding the element of array a if 
        // previous element is from array b 
        dp[i][0] = min(dp[i][0], 
                    dp[i - 1][1] + a[i] + c); 
  
        // Adding the element of array b if 
        // previous element is from array a 
        // with an extra penalty of integer C 
        dp[i][1] = min(dp[i][1], 
                    dp[i - 1][0] + b[i] + c); 
  
        // Adding the element of array b if 
        // previous element is also from array b 
        dp[i][1] = min(dp[i][1], 
                    dp[i - 1][1] + b[i]); 
    } 
  
    // Print the minimum sum 
    cout << min(dp[n - 1][0], 
                dp[n - 1][1]) 
        << "\n"; 
} 
  
// Driver Code 
int main() 
{ 
    // Given array arr1[] and arr2[] 
    int arr1[] = { 7, 6, 18, 6, 16, 
                18, 1, 17, 17 }; 
  
    int arr2[] = { 6, 9, 3, 10, 9, 
                1, 10, 1, 5 }; 
  
    // Given cost 
    int C = 2; 
  
    int N = sizeof(arr1) / sizeof(arr1[0]); 
  
    // Function Call 
    minimumSum(arr1, arr2, C, N); 
  
    return 0; 
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function that prints minimum sum
// after selecting N elements
static void minimumSum(int a[], int b[],
                       int c, int n)
{
      
    // Initialise the dp array
    int [][]dp = new int[n][2];
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            dp[i][j] = (int) 1e6;
        }
    }
      
    // Base Case
    dp[0][0] = a[0];
    dp[0][1] = b[0];
  
    for(int i = 1; i < n; i++) 
    {
          
        // Adding the element of array a if
        // previous element is also from array a
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][0] + a[i]);
  
        // Adding the element of array a if
        // previous element is from array b
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][1] + a[i] + c);
  
        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][0] + b[i] + c);
  
        // Adding the element of array b if
        // previous element is also from array b
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][1] + b[i]);
    }
  
    // Print the minimum sum
    System.out.print(Math.min(dp[n - 1][0],
                              dp[n - 1][1]) + "\n");
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array arr1[] and arr2[]
    int arr1[] = { 7, 6, 18, 6, 16,
                   18, 1, 17, 17 };
  
    int arr2[] = { 6, 9, 3, 10, 9,
                   1, 10, 1, 5 };
  
    // Given cost
    int C = 2;
  
    int N = arr1.length;
  
    // Function call
    minimumSum(arr1, arr2, C, N);
}
}
  
// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach 
  
# Function that prints minimum sum 
# after selecting N elements 
def minimumSum(a, b, c, n): 
      
    # Initialise the dp array 
    dp = [[1e6 for i in range(2)] 
            for j in range(n)] 
  
    # Base Case 
    dp[0][0] = a[0] 
    dp[0][1] = b[0] 
  
    for i in range(1, n): 
  
        # Adding the element of array a if 
        # previous element is also from array a 
        dp[i][0] = min(dp[i][0], 
                    dp[i - 1][0] + a[i]) 
  
        # Adding the element of array a if 
        # previous element is from array b 
        dp[i][0] = min(dp[i][0], 
                    dp[i - 1][1] + a[i] + c) 
  
        # Adding the element of array b if 
        # previous element is from array a 
        # with an extra penalty of integer C 
        dp[i][1] = min(dp[i][1], 
                    dp[i - 1][0] + b[i] + c) 
  
        # Adding the element of array b if 
        # previous element is also from array b 
        dp[i][1] = min(dp[i][1], 
                    dp[i - 1][1] + b[i]) 
  
    # Print the minimum sum 
    print(min(dp[n - 1][0], dp[n - 1][1])) 
  
# Driver code 
if __name__ == '__main__': 
  
    # Given array arr[] 
    arr1 = [ 7, 6, 18, 6, 16, 
            18, 1, 17, 17 ] 
  
    arr2 = [ 6, 9, 3, 10, 9, 
            1, 10, 1, 5 ] 
  
    # Given cost 
    C = 2
  
    N = len(arr1) 
  
    # Function Call 
    minimumSum(arr1, arr2, C, N) 
  
# This code is contributed by Shivam Singh

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function that prints minimum sum
// after selecting N elements
static void minimumSum(int []a, int []b,
                       int c, int n)
{
      
    // Initialise the dp array
    int [,]dp = new int[n, 2];
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            dp[i, j] = (int)1e6;
        }
    }
      
    // Base Case
    dp[0, 0] = a[0];
    dp[0, 1] = b[0];
  
    for(int i = 1; i < n; i++) 
    {
          
        // Adding the element of array a if
        // previous element is also from array a
        dp[i, 0] = Math.Min(dp[i, 0],
                            dp[i - 1, 0] + a[i]);
  
        // Adding the element of array a if
        // previous element is from array b
        dp[i, 0] = Math.Min(dp[i, 0],
                            dp[i - 1, 1] + a[i] + c);
  
        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i, 1] = Math.Min(dp[i, 1],
                            dp[i - 1, 0] + b[i] + c);
  
        // Adding the element of array b if
        // previous element is also from array b
        dp[i, 1] = Math.Min(dp[i, 1],
                            dp[i - 1, 1] + b[i]);
    }
  
    // Print the minimum sum
    Console.Write(Math.Min(dp[n - 1, 0],
                           dp[n - 1, 1]) + "\n");
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array arr1[] and arr2[]
    int []arr1 = { 7, 6, 18, 6, 16,
                   18, 1, 17, 17 };
  
    int []arr2 = { 6, 9, 3, 10, 9,
                   1, 10, 1, 5 };
  
    // Given cost
    int C = 2;
  
    int N = arr1.Length;
  
    // Function call
    minimumSum(arr1, arr2, C, N);
}
}
  
// This code is contributed by Rajput-Ji

输出：

49

时间复杂度： O(N)