从给定的三个数组中选择元素后最小化总和

给定三个相同大小 N 的数组A[] 、 B[]和C[] 。任务是在从这些数组中选择 N 个元素后最小化总和，使得在每个索引i处都有一个来自数组A[i 中任何一个的元素] 、 B[i]或C[i]可以选择，并且不能从同一个数组中选择两个连续的元素。
例子：

Input: A[] = {1, 100, 1}, B[] = {100, 100, 100}, C[] = {100, 100, 100} 
Output: 102 
A[0] + B[1] + A[2] = 1 + 100 + 100 = 201 
A[0] + B[1] + C[2] = 1 + 100 + 100 = 201 
A[0] + C[1] + B[2] = 1 + 100 + 100 = 201 
A[0] + C[1] + A[2] = 1 + 100 + 1 = 102 
B[0] + A[1] + B[2] = 100 + 100 + 100 = 300 
B[0] + A[1] + C[2] = 100 + 100 + 100 = 300 
B[0] + C[1] + A[2] = 100 + 100 + 1 = 201 
B[0] + C[1] + B[2] = 100 + 100 + 100 = 300 
C[0] + A[1] + B[2] = 100 + 100 + 100 = 300 
C[0] + A[1] + C[2] = 100 + 100 + 100 = 300 
C[0] + B[1] + A[2] = 100 + 100 + 1 = 201 
C[0] + B[1] + C[2] = 100 + 100 + 100 = 300
Input: A[] = {1, 1, 1}, B[] = {1, 1, 1}, C[] = {1, 1, 1} 
Output: 3 
 

方法：问题是寻找最小成本的一个简单变体。额外的限制是，如果我们从特定数组中取出一个元素，那么我们不能从同一个数组中取出下一个元素。这可以使用递归轻松解决，但它的时间复杂度为O(3^n)，因为对于每个元素，我们都有三个数组作为选择。
为了提高时间复杂度，我们可以轻松地将预先计算的值存储在 dp 数组中。
由于每个索引都有三个数组可供选择，因此在这种情况下会出现三种情况：

• 情况 1：如果从第 i 个元素中选择数组 A[]，那么我们为第 (i + 1) 个元素选择数组 B[] 或数组 C[]。
• 情况 2：如果从第 i 个元素中选择数组 B[]，那么我们为第 (i + 1) 个元素选择数组 A[] 或数组 C[]。
• 情况 3：如果从第 i 个元素中选择数组 C[]，那么我们为第 (i + 1) 个元素选择数组 A[] 或数组 B[]。

上述状态可以使用递归解决，中间结果可以存储在 dp 数组中。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
 
#include 
using namespace std;
#define SIZE 3
const int N = 3;
 
// Function to return the minimized sum
int minSum(int A[], int B[], int C[], int i,
        int n, int curr, int dp[SIZE][N])
{
 
    // If all the indices have been used
    if (n <= 0)
        return 0;
 
    // If this value is pre-calculated
    // then return its value from dp array
    // instead of re-computing it
    if (dp[n][curr] != -1)
        return dp[n][curr];
 
    // Here curr is the array chosen
    // for the (i - 1)th element
    // 0 for A[], 1 for B[] and 2 for C[]
 
    // If A[i - 1] was chosen previously then
    // only B[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 0) {
        return dp[n][curr]
                = min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp),
                      C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
    }
 
    // If B[i - 1] was chosen previously then
    // only A[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 1)
        return dp[n][curr]
                = min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                      C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
 
    // If C[i - 1] was chosen previously then
    // only A[i] or B[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    return dp[n][curr]
                = min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                      B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp));
}
 
// Driver code
int main()
{
    int A[] = { 1, 50, 1 };
    int B[] = { 50, 50, 50 };
    int C[] = { 50, 50, 50 };
 
    // Initialize the dp[][] array
    int dp[SIZE][N];
    for (int i = 0; i < SIZE; i++)
        for (int j = 0; j < N; j++)
            dp[i][j] = -1;
 
    // min(start with A[0], start with B[0], start with C[0])
    cout << min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp),
                min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp),
                    C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp)));
 
    return 0;
}

Java

// Java implementation of the above approach
import java.io.*;
 
class GFG
{
 
static int SIZE = 3;
static int N = 3;
 
// Function to return the minimized sum
static int minSum(int A[], int B[], int C[], int i,
                    int n, int curr, int [][]dp)
{
 
    // If all the indices have been used
    if (n <= 0)
        return 0;
 
    // If this value is pre-calculated
    // then return its value from dp array
    // instead of re-computing it
    if (dp[n][curr] != -1)
        return dp[n][curr];
 
    // Here curr is the array chosen
    // for the (i - 1)th element
    // 0 for A[], 1 for B[] and 2 for C[]
 
    // If A[i - 1] was chosen previously then
    // only B[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 0)
    {
        return dp[n][curr]
                = Math.min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp),
                    C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
    }
 
    // If B[i - 1] was chosen previously then
    // only A[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 1)
        return dp[n][curr]
                = Math.min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                    C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
 
    // If C[i - 1] was chosen previously then
    // only A[i] or B[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    return dp[n][curr]
                = Math.min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                    B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp));
}
 
// Driver code
public static void main (String[] args)
{
    int A[] = { 1, 50, 1 };
    int B[] = { 50, 50, 50 };
    int C[] = { 50, 50, 50 };
     
    // Initialize the dp[][] array
    int dp[][] = new int[SIZE][N];
    for (int i = 0; i < SIZE; i++)
        for (int j = 0; j < N; j++)
            dp[i][j] = -1;
     
    // min(start with A[0], start with B[0], start with C[0])
    System.out.println(Math.min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp),
                Math.min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp),
                    C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp))));
}
}
 
// This code is contributed by anuj_67..

Python3

# Python3 implementation of the above approach
 
import numpy as np
 
SIZE = 3;
N = 3;
 
# Function to return the minimized sum
def minSum(A, B, C, i, n, curr, dp) :
 
    # If all the indices have been used
    if (n <= 0) :
        return 0;
 
    # If this value is pre-calculated
    # then return its value from dp array
    # instead of re-computing it
    if (dp[n][curr] != -1) :
        return dp[n][curr];
 
    # Here curr is the array chosen
    # for the (i - 1)th element
    # 0 for A[], 1 for B[] and 2 for C[]
 
    # If A[i - 1] was chosen previously then
    # only B[i] or C[i] can chosen now
    # choose the one which leads
    # to the minimum sum
    if (curr == 0) :
        dp[n][curr] = min( B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp),
        C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
        return dp[n][curr]
     
 
    # If B[i - 1] was chosen previously then
    # only A[i] or C[i] can chosen now
    # choose the one which leads
    # to the minimum sum
    if (curr == 1) :
        dp[n][curr] = min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
        C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
        return dp[n][curr]
 
    # If C[i - 1] was chosen previously then
    # only A[i] or B[i] can chosen now
    # choose the one which leads
    # to the minimum sum
    dp[n][curr] = min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
    B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp));
     
    return dp[n][curr]
 
 
# Driver code
if __name__ == "__main__" :
 
    A = [ 1, 50, 1 ];
    B = [ 50, 50, 50 ];
    C = [ 50, 50, 50 ];
 
    # Initialize the dp[][] array
    dp = np.zeros((SIZE,N));
     
    for i in range(SIZE) :
        for j in range(N) :
            dp[i][j] = -1;
 
    # min(start with A[0], start with B[0], start with C[0])
    print(min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp),
                min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp),
                C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp))));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
static int SIZE = 3;
static int N = 3;
 
// Function to return the minimized sum
static int minSum(int []A, int []B, int []C, int i,
                    int n, int curr, int [,]dp)
{
 
    // If all the indices have been used
    if (n <= 0)
        return 0;
 
    // If this value is pre-calculated
    // then return its value from dp array
    // instead of re-computing it
    if (dp[n,curr] != -1)
        return dp[n,curr];
 
    // Here curr is the array chosen
    // for the (i - 1)th element
    // 0 for A[], 1 for B[] and 2 for C[]
 
    // If A[i - 1] was chosen previously then
    // only B[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 0)
    {
        return dp[n,curr]
                = Math.Min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp),
                    C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
    }
 
    // If B[i - 1] was chosen previously then
    // only A[i] or C[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    if (curr == 1)
        return dp[n,curr]
                = Math.Min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                    C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
 
    // If C[i - 1] was chosen previously then
    // only A[i] or B[i] can chosen now
    // choose the one which leads
    // to the minimum sum
    return dp[n,curr]
                = Math.Min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                    B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp));
}
 
// Driver code
public static void Main ()
{
    int []A = { 1, 50, 1 };
    int []B = { 50, 50, 50 };
    int []C = { 50, 50, 50 };
     
    // Initialize the dp[][] array
    int [,]dp = new int[SIZE,N];
    for (int i = 0; i < SIZE; i++)
        for (int j = 0; j < N; j++)
            dp[i,j] = -1;
     
    // min(start with A[0], start with B[0], start with C[0])
    Console.WriteLine(Math.Min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp),
                Math.Min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp),
                    C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp))));
}
}
 
// This code is contributed by anuj_67..

Javascript

    // Javascript implementation of the above approach
     
    let SIZE = 3;
    let N = 3;
 
    // Function to return the minimized sum
    function minSum(A, B, C, i, n, curr, dp)
    {
 
        // If all the indices have been used
        if (n <= 0)
            return 0;
 
        // If this value is pre-calculated
        // then return its value from dp array
        // instead of re-computing it
        if (dp[n][curr] != -1)
            return dp[n][curr];
 
        // Here curr is the array chosen
        // for the (i - 1)th element
        // 0 for A[], 1 for B[] and 2 for C[]
 
        // If A[i - 1] was chosen previously then
        // only B[i] or C[i] can chosen now
        // choose the one which leads
        // to the minimum sum
        if (curr == 0)
        {
            return dp[n][curr]
                    = Math.min(B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp),
                        C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
        }
 
        // If B[i - 1] was chosen previously then
        // only A[i] or C[i] can chosen now
        // choose the one which leads
        // to the minimum sum
        if (curr == 1)
            return dp[n][curr]
                    = Math.min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                        C[i] + minSum(A, B, C, i + 1, n - 1, 2, dp));
 
        // If C[i - 1] was chosen previously then
        // only A[i] or B[i] can chosen now
        // choose the one which leads
        // to the minimum sum
        return dp[n][curr]
                    = Math.min(A[i] + minSum(A, B, C, i + 1, n - 1, 0, dp),
                        B[i] + minSum(A, B, C, i + 1, n - 1, 1, dp));
    }
     
    let A = [ 1, 50, 1 ];
    let B = [ 50, 50, 50 ];
    let C = [ 50, 50, 50 ];
       
    // Initialize the dp[][] array
    let dp = new Array(SIZE);
    for (let i = 0; i < SIZE; i++)
    {
        dp[i] = new Array(N);
        for (let j = 0; j < N; j++)
        {
            dp[i][j] = -1;
        }
    }
       
    // min(start with A[0], start with B[0], start with C[0])
    document.write(Math.min(A[0] + minSum(A, B, C, 1, SIZE - 1, 0, dp),
                Math.min(B[0] + minSum(A, B, C, 1, SIZE - 1, 1, dp),
                    C[0] + minSum(A, B, C, 1, SIZE - 1, 2, dp))));
 

输出：

52

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