我和乘子交替方向法admm_找到最大和交替子序列

我和乘子交替方向法admm

Problem statement:

问题陈述：

Given a sequence of numbers, you have to find the maximum sum alternating subsequence and print the value. A sequence is an alternating sequence when it will be maintain like (increasing) -> (decreasing) ->(increasing) ->(decreasing).

给定一个数字序列，您必须找到最大和交替子序列并打印值 。当序列将像(增加)->(减少)->(增加)->(减少)一样被维护时，它是一个交替的序列。

Input:
T Test case
T no. of input string will be given to you.
E.g.
3
2 3 4 8 2 5 6 8
2 3 4 8 2 6 5 4
6 5 9 2 10 77 5
Constrain:
1≤ A[i] ≤50
Output:
Print the value of maximum sum alternating subsequence.

Example

例

T=3
Input:
2 3 4 8 2 5 6 8 
Output:
22 ( 8+6+8)
Input:
2 3 4 8 2 6 5 4
Output:
20 ( 8+ 2+ 6+ 4)
Input:
6 5 9 2 10 77 5
Output:
98 (5+ 9+ 2+ 77+5)

Explanation with example:

举例说明：

Let N be the number of elements say, X1, X2, X3, ..., Xn

令N为元素的数量，即X1，X2，X3，...，Xn

Let f(a) = the value at the index a of the increasing array, and g(a) = the value at the index a of the decreasing array.

令f(a) =递增数组的索引a处的值， g(a) =递减数组的索引a处的值。

To find out the maximum sum alternating sequence we will follow these steps,

为了找出最大和交替序列，我们将按照以下步骤操作，

We take two new arrays, one is increasing array and another is decreasing array and initialize it with 0. We start our algorithm with the second column. We check elements that are before the current element, with the current element.
我们采用两个新数组，一个是递增数组，另一个是递减数组，并将其初始化为0。我们从第二列开始我们的算法。我们使用当前元素检查当前元素之前的元素。
If any element is less than the current element then,
如果任何元素小于当前元素，

f(indexofthecurrentelement) = max⁡
f(当前元素的索引)=max⁡
If the element is greater than the current element then,
如果元素大于当前元素，

g(indexofthecurrentelement) = max⁡
g(当前元素的索引)=max⁡

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int sum(int* arr, int n)
{
int inc[n + 1], dec[n + 1];
inc[0] = arr[0];
dec[0] = arr[0];
memset(inc, 0, sizeof(inc));
memset(dec, 0, sizeof(dec));
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (arr[j] > arr[i]) {
dec[i] = max(dec[i], inc[j] + arr[i]);
}
else if (arr[i] > arr[j]) {
inc[i] = max(inc[i], dec[j] + arr[i]);
}
}
}
return max(inc[n - 1], dec[n - 1]);
}
int main()
{
int t;
cout << "Test Case : ";
cin >> t;
while (t--) {
int n;
cout << "Number of element : ";
cin >> n;
int arr[n];
cout << "Enter the elements : ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
cout << "Sum of the alternating sequence : " << sum(arr, n) << endl;
}
return 0;
}

Output

输出量

Test Case : 3
Number of element : 8
Enter the elements : 2 3 4 8 2 5 6 8
Sum of the alternating sequence : 22
Number of element : 8              
Enter the elements : 2 3 4 8 2 6 5 4
Sum of the alternating sequence : 20
Number of element : 7
Enter the elements : 6 5 9 2 10 77 5
Sum of the alternating sequence : 98