合并 List 的所有元素的最低成本

给定一个包含N 个整数的列表，任务是将列表中的所有元素以可能的最小成本合并为一个。合并规则如下：
选择列表中任意两个相邻元素，其值为X和Y ，并将它们合并为一个值为(X + Y)的单个元素，总成本为(X + Y) 。
例子：

Input: arr[] = {1, 3, 7} 
Output: 15 
All possible ways of merging: 
a) {1, 3, 7} (cost = 0) -> {4, 7} (cost = 0+4 = 4) -> 11 (cost = 4+11 = 15) 
b) {1, 3, 7} (cost = 0) -> {1, 10} (cost = 0+10= 10) -> 11 (cost = 10+11 = 21) 
Thus, ans = 15
Input: arr[] = {1, 2, 3, 4} 
Output: 19 
 

方法：这个问题可以用动态规划解决。让我们首先定义 DP 的状态。 DP[l][r]将是将子数组 arr[l…r]合并为一个的最小成本。
现在，让我们看一下递归关系：

DP[l][r] = min(S(l, l) + S(l + 1, r) + DP[l][l] + DP[l + 1][r], S(l, l + 1) + S(l + 2, r) + DP[l][l + 1] + DP[l + 2][r], …, S(l, r – 1) + S(r, r) + DP[l][r – 1] + DP[r][r]) = S(l, r) + min(DP[l][l] + DP[l + 1][r], DP[l][l + 1] + DP[l + 2][r], …, DP[l][r – 1] + DP[r][r]) 
where S(x, y) is the sum of all the elements of the subarray arr[x…y] 
 

该方法的时间复杂度为O(N ³ )，因为总共有N * N个状态，并且每个状态通常需要O(N)时间来解决。
下面是上述方法的实现：

CPP

// C++ implementation of the approach
#include 
using namespace std;
 
#define N 401
 
// To store the states of DP
int dp[N][N];
bool v[N][N];
 
// Function to return the minimum merge cost
int minMergeCost(int i, int j, int* arr)
{
 
    // Base case
    if (i == j)
        return 0;
 
    // If the state has been solved before
    if (v[i][j])
        return dp[i][j];
 
    // Marking the state as solved
    v[i][j] = 1;
    int& x = dp[i][j];
 
    // Reference to dp[i][j]
    x = INT_MAX;
 
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
 
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++) {
        x = min(x, tot + minMergeCost(i, k - 1, arr)
                       + minMergeCost(k, j, arr));
    }
 
    // Returning the solved value
    return x;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 3, 7 };
    int n = sizeof(arr) / sizeof(int);
 
    cout << minMergeCost(0, n - 1, arr);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
static final int N = 401;
 
// To store the states of DP
static int [][]dp = new int[N][N];
static boolean [][]v = new boolean[N][N];
 
// Function to return the minimum merge cost
static int minMergeCost(int i, int j, int[] arr)
{
 
    // Base case
    if (i == j)
        return 0;
 
    // If the state has been solved before
    if (v[i][j])
        return dp[i][j];
 
    // Marking the state as solved
    v[i][j] = true;
    int x = dp[i][j];
 
    // Reference to dp[i][j]
    x = Integer.MAX_VALUE;
 
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
 
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++)
    {
        x = Math.min(x, tot + minMergeCost(i, k - 1, arr)
                    + minMergeCost(k, j, arr));
    }
 
    // Returning the solved value
    return x;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 3, 7 };
    int n = arr.length;
 
    System.out.print(minMergeCost(0, n - 1, arr));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
import sys
 
N = 401;
 
# To store the states of DP
dp = [[0 for i in range(N)] for j in range(N)];
v = [[False for i in range(N)] for j in range(N)];
 
# Function to return the minimum merge cost
def minMergeCost(i, j, arr):
 
    # Base case
    if (i == j):
        return 0;
 
    # If the state has been solved before
    if (v[i][j]):
        return dp[i][j];
 
    # Marking the state as solved
    v[i][j] = True;
    x = dp[i][j];
 
    # Reference to dp[i][j]
    x = sys.maxsize;
 
    # To store the sum of all the
    # elements in the subarray arr[i...j]
    tot = 0;
    for k in range(i, j + 1):
        tot += arr[k];
 
    # Loop to iterate the recurrence
    for k in range(i + 1, j + 1):
        x = min(x, tot + minMergeCost(i, k - 1, arr) + \
                minMergeCost(k, j, arr));
     
    # Returning the solved value
    return x;
 
# Driver code
if __name__ == '__main__':
    arr = [ 1, 3, 7 ];
    n = len(arr);
 
    print(minMergeCost(0, n - 1, arr));
 
# This code is contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
static readonly int N = 401;
 
// To store the states of DP
static int [,]dp = new int[N, N];
static bool [,]v = new bool[N, N];
 
// Function to return the minimum merge cost
static int minMergeCost(int i, int j, int[] arr)
{
 
    // Base case
    if (i == j)
        return 0;
 
    // If the state has been solved before
    if (v[i, j])
        return dp[i, j];
 
    // Marking the state as solved
    v[i, j] = true;
    int x = dp[i, j];
 
    // Reference to dp[i,j]
    x = int.MaxValue;
 
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    int tot = 0;
    for (int k = i; k <= j; k++)
        tot += arr[k];
 
    // Loop to iterate the recurrence
    for (int k = i + 1; k <= j; k++)
    {
        x = Math.Min(x, tot + minMergeCost(i, k - 1, arr)
                    + minMergeCost(k, j, arr));
    }
 
    // Returning the solved value
    return x;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 3, 7 };
    int n = arr.Length;
 
    Console.Write(minMergeCost(0, n - 1, arr));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
var N = 401
 
// To store the states of DP
var dp = Array.from(Array(N), ()=> Array(N));
var v = Array.from(Array(N), ()=> Array(N));
 
// Function to return the minimum merge cost
function minMergeCost(i, j, arr)
{
 
    // Base case
    if (i == j)
        return 0;
 
    // If the state has been solved before
    if (v[i][j])
        return dp[i][j];
 
    // Marking the state as solved
    v[i][j] = 1;
    var x = dp[i][j];
 
    // Reference to dp[i][j]
    x = 1000000000;
 
    // To store the sum of all the
    // elements in the subarray arr[i...j]
    var tot = 0;
    for (var k = i; k <= j; k++)
        tot += arr[k];
 
    // Loop to iterate the recurrence
    for (var k = i + 1; k <= j; k++) {
        x = Math.min(x, tot + minMergeCost(i, k - 1, arr)
                       + minMergeCost(k, j, arr));
    }
 
    // Returning the solved value
    return x;
}
 
// Driver code
var arr = [1, 3, 7];
var n = arr.length;
document.write( minMergeCost(0, n - 1, arr));
 

输出：

15

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