给定N个元素，您可以从列表中删除任何两个元素，记下它们的总和，然后将总和添加到列表中。当列表中有多个元素时，请重复这些步骤。最后的任务是使这些选定总和的总和最小化。
例子：
Input: arr[] = {1, 4, 7, 10}
Output: 39
Choose 1 and 4, Sum = 5, arr[] = {5, 7, 10}
Choose 5 and 7, Sum = 17, arr[] = {12, 10}
Choose 12 and 10, Sum = 39, arr[] = {22}
Input: arr[] = {1, 3, 7, 5, 6}
Output: 48
方法：为了使总和最小，在每个步骤中选择的元素必须是列表中的最小元素。为了有效地做到这一点，可以使用优先级队列。在每个步骤中，当列表中有多个元素时，请选择最小值和第二个最小值，从列表中删除它们，并在更新运行总和后将其总和添加到列表中。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the minimized sum
int getMinSum(int arr[], int n)
{
int i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
priority_queue, greater > pq;
// Add all the elements
// to the priority queue
for (i = 0; i < n; i++)
pq.push(arr[i]);
// While there are more than 1 elements
// left in the queue
while (pq.size() > 1)
{
// Remove and get the minimum
// element from the queue
int min = pq.top();
pq.pop();
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq.top();
pq.pop();
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.push(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
int main()
{
int arr[] = { 1, 3, 7, 5, 6 };
int n = sizeof(arr)/sizeof(arr[0]);
cout << (getMinSum(arr, n));
}
// This code is contributed by mohit Java
// Java implementation of the approach
import java.util.PriorityQueue;
class GFG
{
// Function to return the minimized sum
static int getMinSum(int arr[], int n)
{
int i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
PriorityQueue pq = new PriorityQueue<>();
// Add all the elements
// to the prioriry queue
for (i = 0; i < n; i++)
pq.add(arr[i]);
// While there are more than 1 elements
// left in the queue
while (pq.size() > 1)
{
// Remove and get the minimum
// element from the queue
int min = pq.poll();
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq.poll();
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.add(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 3, 7, 5, 6 };
int n = arr.length;
System.out.print(getMinSum(arr, n));
}
} Python3
# Python3 implementation of the approach
# Function to return the minimized sum
def getMinSum(arr, n):
sum = 0
# Priority queue to store the elements of the array
# and retrieve the minimum element efficiently
pq = []
# Add all the elements
# to the priority queue
for i in range( n ):
pq.append(arr[i])
# While there are more than 1 elements
# left in the queue
while (len(pq) > 1) :
pq.sort(reverse=True)
# Remove and get the minimum
# element from the queue
min = pq[-1];
pq.pop();
# Remove and get the second minimum
# element (currently minimum)
secondMin = pq[-1];
pq.pop();
# Update the sum
sum += (min + secondMin);
# Add the sum of the minimum
# elements to the queue
pq.append(min + secondMin)
# Return the minimized sum
return sum
# Driver code
if __name__ == "__main__":
arr = [ 1, 3, 7, 5, 6 ]
n = len(arr)
print(getMinSum(arr, n))
# This code is contributed by chitranayalC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the minimized sum
static int getMinSum(int[] arr, int n)
{
int i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
List pq = new List();
// Add all the elements
// to the priority queue
for (i = 0; i < n; i++)
{
pq.Add(arr[i]);
}
// While there are more than 1 elements
// left in the queue
while(pq.Count > 1)
{
pq.Sort();
// Remove and get the minimum
// element from the queue
int min = pq[0];
pq.RemoveAt(0);
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq[0];
pq.RemoveAt(0);
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.Add(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
static public void Main ()
{
int[] arr = { 1, 3, 7, 5, 6 };
int n = arr.Length;
Console.WriteLine(getMinSum(arr, n));
}
}
// This code is contributed by avanitrachhadiya2155 输出：
48