给定三个由N 个整数组成的数组A[] 、 B[]和C[] 。我们可以从这个数组中选择N 个元素，这样对于每个索引i只能从这些数组中选择一个元素，即A[i] 、 B[i]或C[i]并且不能从相同的元素中选择两个连续的元素大批。任务是通过从这些数组中选择元素来打印我们可以得出的最大数字总和。
例子：
Input: a[] = {10, 20, 30}, b[] = {40, 50, 60}, c[] = {70, 80, 90}
Output: 210
70 + 50 + 90 = 210
Input: a[] = {6, 8, 2, 7, 4, 2, 7}, b[] = {7, 8, 5, 8, 6, 3, 5}, c[] = {8, 3, 2, 6, 8, 4, 1}
Output: 46
Choose elements from C, A, B, A, C, B and A.
方法：上述问题可以使用动态规划解决。如果选择第 j 个数组中的元素，则将dp[i][j]视为直到 i 的最大和。我们可以从任何数组中为第一个索引选择元素，但稍后递归地我们只能从其余两个数组中选择一个元素用于下一步。所有组合返回的最大总和将是我们的答案。使用记忆化来避免重复和多次相同的函数调用。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
const int N = 3;
// Function to return the maximum sum
int FindMaximumSum(int ind, int kon, int a[], int b[],
int c[], int n, int dp[][N])
{
// Base case
if (ind == n)
return 0;
// Already visited
if (dp[ind][kon] != -1)
return dp[ind][kon];
int ans = -1e9 + 5;
// If the element has been taken
// from first array in previous step
if (kon == 0) {
ans = max(ans, b[ind] + FindMaximumSum(ind + 1,
1, a, b,
c, n, dp));
ans = max(ans, c[ind] + FindMaximumSum(ind + 1,
2, a, b,
c, n, dp));
}
// If the element has been taken
// from second array in previous step
else if (kon == 1) {
ans = max(ans, a[ind] + FindMaximumSum(ind + 1,
0, a, b,
c, n, dp));
ans = max(ans, c[ind] + FindMaximumSum(ind + 1,
2, a, b,
c, n, dp));
}
// If the element has been taken
// from third array in previous step
else if (kon == 2) {
ans = max(ans, a[ind] + FindMaximumSum(ind + 1,
1, a, b,
c, n, dp));
ans = max(ans, b[ind] + FindMaximumSum(ind + 1,
0, a, b,
c, n, dp));
}
return dp[ind][kon] = ans;
}
// Driver code
int main()
{
int a[] = { 6, 8, 2, 7, 4, 2, 7 };
int b[] = { 7, 8, 5, 8, 6, 3, 5 };
int c[] = { 8, 3, 2, 6, 8, 4, 1 };
int n = sizeof(a) / sizeof(a[0]);
int dp[n][N];
memset(dp, -1, sizeof dp);
// Pick element from first array
int x = FindMaximumSum(0, 0, a, b, c, n, dp);
// Pick element from second array
int y = FindMaximumSum(0, 1, a, b, c, n, dp);
// Pick element from third array
int z = FindMaximumSum(0, 2, a, b, c, n, dp);
// Print the maximum of them
cout << max(x, max(y, z));
return 0;
} Java
// Java program for the above approach
class GFG {
static int N = 3;
// Function to return the maximum sum
static int FindMaximumSum(int ind, int kon, int a[],
int b[], int c[], int n, int dp[][])
{
// Base case
if (ind == n)
return 0;
// Already visited
if (dp[ind][kon] != -1)
return dp[ind][kon];
int ans = (int) (-1e9 + 5);
// If the element has been taken
// from first array in previous step
if (kon == 0) {
ans = Math.max(ans, b[ind] +
FindMaximumSum(ind + 1,
1, a, b,c, n, dp));
ans = Math.max(ans, c[ind] +
FindMaximumSum(ind + 1,
2, a, b,c, n, dp));
}
// If the element has been taken
// from second array in previous step
else if (kon == 1) {
ans = Math.max(ans, a[ind] +
FindMaximumSum(ind + 1,
0, a, b, c, n, dp));
ans = Math.max(ans, c[ind] +
FindMaximumSum(ind + 1,
2, a, b, c, n, dp));
}
// If the element has been taken
// from third array in previous step
else if (kon == 2) {
ans = Math.max(ans, a[ind] +
FindMaximumSum(ind + 1,
1, a, b, c, n, dp));
ans = Math.max(ans, b[ind] +
FindMaximumSum(ind + 1,
0, a, b, c, n, dp));
}
return dp[ind][kon] = ans;
}
// Driver code
public static void main(String[] args) {
int a[] = { 6, 8, 2, 7, 4, 2, 7 };
int b[] = { 7, 8, 5, 8, 6, 3, 5 };
int c[] = { 8, 3, 2, 6, 8, 4, 1 };
int n = a.length;
int dp[][] = new int[n][N];
for(int i = 0; i < n; i++) {
for(int j = 0; j < N; j++) {
dp[i][j] =- 1;
}
}
// Pick element from first array
int x = FindMaximumSum(0, 0, a, b, c, n, dp);
// Pick element from second array
int y = FindMaximumSum(0, 1, a, b, c, n, dp);
// Pick element from third array
int z = FindMaximumSum(0, 2, a, b, c, n, dp);
// Print the maximum of them
System.out.println(Math.max(x, Math.max(y, z)));
}
}
// This code has been contributed
// by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to return the maximum sum
def FindMaximumSum(ind, kon, a, b, c, n, dp):
# Base case
if ind == n:
return 0
# Already visited
if dp[ind][kon] != -1:
return dp[ind][kon]
ans = -10 ** 9 + 5
# If the element has been taken
# from first array in previous step
if kon == 0:
ans = max(ans, b[ind] +
FindMaximumSum(ind + 1, 1,
a, b, c, n, dp))
ans = max(ans, c[ind] +
FindMaximumSum(ind + 1, 2,
a, b, c, n, dp))
# If the element has been taken
# from second array in previous step
elif kon == 1:
ans = max(ans, a[ind] +
FindMaximumSum(ind + 1, 0,
a, b, c, n, dp))
ans = max(ans, c[ind] +
FindMaximumSum(ind + 1, 2,
a, b, c, n, dp))
# If the element has been taken
# from third array in previous step
elif kon == 2:
ans = max(ans, a[ind] +
FindMaximumSum(ind + 1, 1,
a, b, c, n, dp))
ans = max(ans, b[ind] +
FindMaximumSum(ind + 1, 0,
a, b, c, n, dp))
dp[ind][kon] = ans
return ans
# Driver code
if __name__ == "__main__":
N = 3
a = [6, 8, 2, 7, 4, 2, 7]
b = [7, 8, 5, 8, 6, 3, 5]
c = [8, 3, 2, 6, 8, 4, 1]
n = len(a)
dp = [[-1 for i in range(N)]
for j in range(n)]
# Pick element from first array
x = FindMaximumSum(0, 0, a, b, c, n, dp)
# Pick element from second array
y = FindMaximumSum(0, 1, a, b, c, n, dp)
# Pick element from third array
z = FindMaximumSum(0, 2, a, b, c, n, dp)
# Print the maximum of them
print(max(x, y, z))
# This code is contributed
# by Rituraj JainC#
// C# program for the above approach
using System;
class GFG
{
static int N = 3;
// Function to return the maximum sum
static int FindMaximumSum(int ind, int kon, int []a,
int []b, int []c, int n, int [,]dp)
{
// Base case
if (ind == n)
return 0;
// Already visited
if (dp[ind,kon] != -1)
return dp[ind,kon];
int ans = (int) (-1e9 + 5);
// If the element has been taken
// from first array in previous step
if (kon == 0)
{
ans = Math.Max(ans, b[ind] +
FindMaximumSum(ind + 1,
1, a, b,c, n, dp));
ans = Math.Max(ans, c[ind] +
FindMaximumSum(ind + 1,
2, a, b,c, n, dp));
}
// If the element has been taken
// from second array in previous step
else if (kon == 1)
{
ans = Math.Max(ans, a[ind] +
FindMaximumSum(ind + 1,
0, a, b, c, n, dp));
ans = Math.Max(ans, c[ind] +
FindMaximumSum(ind + 1,
2, a, b, c, n, dp));
}
// If the element has been taken
// from third array in previous step
else if (kon == 2)
{
ans = Math.Max(ans, a[ind] +
FindMaximumSum(ind + 1,
1, a, b, c, n, dp));
ans = Math.Max(ans, b[ind] +
FindMaximumSum(ind + 1,
0, a, b, c, n, dp));
}
return dp[ind,kon] = ans;
}
// Driver code
public static void Main()
{
int []a = { 6, 8, 2, 7, 4, 2, 7 };
int []b = { 7, 8, 5, 8, 6, 3, 5 };
int []c = { 8, 3, 2, 6, 8, 4, 1 };
int n = a.Length;
int [,]dp = new int[n,N];
for(int i = 0; i < n; i++)
{
for(int j = 0; j < N; j++)
{
dp[i,j] =- 1;
}
}
// Pick element from first array
int x = FindMaximumSum(0, 0, a, b, c, n, dp);
// Pick element from second array
int y = FindMaximumSum(0, 1, a, b, c, n, dp);
// Pick element from third array
int z = FindMaximumSum(0, 2, a, b, c, n, dp);
// Print the maximum of them
Console.WriteLine(Math.Max(x, Math.Max(y, z)));
}
}
// This code has been contributed by RyugaPHP
Javascript
输出：
45时间复杂度： O(N)
辅助空间： O(N)
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