矩阵链乘法 | DP-8

给定一个矩阵序列，找到将这些矩阵相乘的最有效方法。问题实际上不是执行乘法，而只是决定执行乘法的顺序。
我们有很多选择来乘以矩阵链，因为矩阵乘法是关联的。换句话说，无论我们如何将乘积括起来，结果都是一样的。例如，如果我们有四个矩阵 A、B、C 和 D，我们将有：

(ABC)D = (AB)(CD) = A(BCD) = ....

但是，我们将乘积括起来的顺序会影响计算乘积所需的简单算术运算的数量或效率。例如，假设 A 是 10 × 30 矩阵，B 是 30 × 5 矩阵，C 是 5 × 60 矩阵。然后，

(AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.

显然，第一个括号需要较少的操作。
给定一个数组 p[]，它表示矩阵链，使得第 i 个矩阵 Ai 的维度为 p[i-1] xp[i]。我们需要编写一个函数MatrixChainOrder()，它应该返回乘法链所需的最小乘法次数。

Input: p[] = {40, 20, 30, 10, 30}   
Output: 26000  
There are 4 matrices of dimensions 40x20, 20x30, 30x10 and 10x30.
Let the input 4 matrices be A, B, C and D.  The minimum number of 
multiplications are obtained by putting parenthesis in following way
(A(BC))D --> 20*30*10 + 40*20*10 + 40*10*30

Input: p[] = {10, 20, 30, 40, 30} 
Output: 30000 
There are 4 matrices of dimensions 10x20, 20x30, 30x40 and 40x30. 
Let the input 4 matrices be A, B, C and D.  The minimum number of 
multiplications are obtained by putting parenthesis in following way
((AB)C)D --> 10*20*30 + 10*30*40 + 10*40*30

Input: p[] = {10, 20, 30}  
Output: 6000  
There are only two matrices of dimensions 10x20 and 20x30. So there 
is only one way to multiply the matrices, cost of which is 10*20*30

1) 最优子结构：
一个简单的解决方案是在所有可能的位置放置括号，计算每个放置的成本并返回最小值。在大小为 n 的矩阵链中，我们可以以 n-1 种方式放置第一组括号。例如，如果给定的链是 4 个矩阵。设链为ABCD，则第一组括号外侧有3种方式：(A)(BCD)、(AB)(CD)和(ABC)(D)。因此，当我们放置一组括号时，我们将问题划分为更小的子问题。因此，该问题具有最优子结构性质，并且可以使用递归轻松解决。
将大小为 n 的链相乘所需的最小乘法次数 = 所有 n-1 个放置的最小值(这些放置创建更小的子问题)

2) 重叠子问题
以下是简单遵循上述最优子结构属性的递归实现。

下面是上述想法的实现：

C++

/* A naive recursive implementation that simply
follows the above optimal substructure property */
#include 
using namespace std;
 
// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1..n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;
 
    // place parenthesis at different places
    // between first and last matrix, recursively
    // calculate count of multiplications for
    // each parenthesis placement and return the
    // minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, 1, n - 1);
}
 
// This code is contributed by Shivi_Aggarwal

C

/* A naive recursive implementation that simply
  follows the above optimal substructure property */
#include 
#include 
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;
 
    // place parenthesis at different places between first
    // and last matrix, recursively calculate count of
    // multiplications for each parenthesis placement and
    // return the minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, 1, n - 1));
 
    getchar();
    return 0;
}

Java

/* A naive recursive implementation that simply follows
   the above optimal substructure property */
class MatrixChainMultiplication {
    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    static int MatrixChainOrder(int p[], int i, int j)
    {
        if (i == j)
            return 0;
 
        int min = Integer.MAX_VALUE;
 
        // place parenthesis at different places between
        // first and last matrix, recursively calculate
        // count of multiplications for each parenthesis
        // placement and return the minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];
 
            if (count < min)
                min = count;
        }
 
        // Return minimum count
        return min;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4, 3 };
        int n = arr.length;
 
        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, n - 1));
    }
}
/* This code is contributed by Rajat Mishra*/

Python3

# A naive recursive implementation that
# simply follows the above optimal
# substructure property
import sys
 
# Matrix A[i] has dimension p[i-1] x p[i]
# for i = 1..n
 
 
def MatrixChainOrder(p, i, j):
 
    if i == j:
        return 0
 
    _min = sys.maxsize
 
    # place parenthesis at different places
    # between first and last matrix,
    # recursively calculate count of
    # multiplications for each parenthesis
    # placement and return the minimum count
    for k in range(i, j):
 
        count = (MatrixChainOrder(p, i, k)
                 + MatrixChainOrder(p, k + 1, j)
                 + p[i-1] * p[k] * p[j])
 
        if count < _min:
            _min = count
 
    # Return minimum count
    return _min
 
 
# Driver code
arr = [1, 2, 3, 4, 3]
n = len(arr)
 
print("Minimum number of multiplications is ",
      MatrixChainOrder(arr, 1, n-1))
 
# This code is contributed by Aryan Garg

C#

/* C# code for naive recursive implementation
that simply follows the above optimal
substructure property */
using System;
 
class GFG {
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int i, int j)
    {
 
        if (i == j)
            return 0;
 
        int min = int.MaxValue;
 
        // place parenthesis at different places
        // between first and last matrix, recursively
        // calculate count of multiplications for each
        // parenthesis placement and return the
        // minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];
 
            if (count < min)
                min = count;
        }
 
        // Return minimum count
        return min;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4, 3 };
        int n = arr.Length;
 
        Console.Write(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, n - 1));
    }
}
 
// This code is contributed by Sam007.

Javascript

 
/* A naive recursive implementation that simply follows
   the above optimal substructure property */
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(p , i , j)
{
    if (i == j)
        return 0;
 
    var min = Number.MAX_VALUE;
 
    // place parenthesis at different places between
    // first and last matrix, recursively calculate
    // count of multiplications for each parenthesis
    // placement and return the minimum count
    var k=0;
    for (k = i; k < j; k++)
    {
        var count = MatrixChainOrder(p, i, k)
                    + MatrixChainOrder(p, k + 1, j)
                    + p[i - 1] * p[k] * p[j];
 
        if (count < min)
            min = count;
    }
 
    // Return minimum count
    return min;
}
 
// Driver code
var arr = [ 1, 2, 3, 4, 3 ];
var n = arr.length;
 
document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, 1, n - 1));
 
// This code contributed by shikhasingrajput
 

输出

Minimum number of multiplications is 30

上述朴素递归方法的时间复杂度是指数级的。需要注意的是，上述函数一次又一次地计算相同的子问题。对于大小为 4 的矩阵链，请参见以下递归树。函数MatrixChainOrder(p, 3, 4) 被调用两次。我们可以看到有很多子问题被多次调用。

由于再次调用相同的子问题，因此该问题具有重叠子问题的属性。因此，矩阵链乘法问题具有动态规划问题的两个属性(参见this和this)。与其他典型的动态规划 (DP) 问题一样，通过以自下而上的方式构造临时数组 m[][] 可以避免相同子问题的重新计算。

动态规划解决方案
以下是使用动态规划(制表 vs 记忆)的矩阵链乘法问题的实现

使用记忆——

C++

// C++ program using memoization
#include 
using namespace std;
int dp[100][100];
 
// Function for matrix chain multiplication
int matrixChainMemoised(int* p, int i, int j)
{
    if (i == j)
    {
        return 0;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
    dp[i][j] = INT_MAX;
    for (int k = i; k < j; k++)
    {
        dp[i][j] = min(
            dp[i][j], matrixChainMemoised(p, i, k)
                     + matrixChainMemoised(p, k + 1, j)
                       + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
int MatrixChainOrder(int* p, int n)
{
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    memset(dp, -1, sizeof dp);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, n);
}
 
// This code is contributed by Sumit_Yadav

Java

// Java program using memoization
import java.io.*;
import java.util.*;
class GFG
{
 
  static int[][] dp = new int[100][100];
 
  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j) 
    {
      return 0;
    }
    if (dp[i][j] != -1) 
    {
      return dp[i][j];
    }
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) 
    {
      dp[i][j] = Math.min(
        dp[i][j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
  }
 
  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
 
    int arr[] = { 1, 2, 3, 4 };
    int n= arr.length;
 
    for (int[] row : dp)
      Arrays.fill(row, -1);
 
    System.out.println("Minimum number of multiplications is " + MatrixChainOrder(arr, n));
  }
}
 
// This code is contributed by avanitrachhadiya2155

蟒蛇3

# Python program using memoization
import sys
dp = [[-1 for i in range(100)] for j in range(100)]
 
# Function for matrix chain multiplication
def matrixChainMemoised(p, i, j):
    if(i == j):
        return 0
     
    if(dp[i][j] != -1):
        return dp[i][j]
     
    dp[i][j] = sys.maxsize
     
    for k in range(i,j):
        dp[i][j] = min(dp[i][j], matrixChainMemoised(p, i, k) + matrixChainMemoised(p, k + 1, j)+ p[i - 1] * p[k] * p[j])
     
    return dp[i][j]
 
def MatrixChainOrder(p,n):
    i = 1
    j = n - 1   
    return matrixChainMemoised(p, i, j)
 
# Driver Code
arr = [1, 2, 3, 4]
n = len(arr)
print("Minimum number of multiplications is",MatrixChainOrder(arr, n))
 
# This code is contributed by rag2127

C＃

// C# program using memoization
using System;
class GFG
{
     
    static int[,] dp = new int[100, 100];
  
  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j) 
    {
      return 0;
    }
    if (dp[i, j] != -1) 
    {
      return dp[i, j];
    }
    dp[i, j] = Int32.MaxValue;
    for (int k = i; k < j; k++) 
    {
      dp[i, j] = Math.Min(
        dp[i, j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i,j];
  }
  
  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }
   
  // Driver code
  static void Main()
  {
    int[] arr = { 1, 2, 3, 4 };
    int n = arr.Length;
     
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
        {
            dp[i, j] = -1;
        }
    }
  
    Console.WriteLine("Minimum number of multiplications is " +
                      MatrixChainOrder(arr, n));
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

 
// Javascript program using memoization
let dp = new Array(100);
for(var i = 0; i < dp.length; i++)
{
    dp[i] = new Array(2);
}
 
// Function for matrix chain multiplication
function matrixChainMemoised(p, i, j)
{
    if (i == j) 
    {
        return 0;
    }
    if (dp[i][j] != -1) 
    {
        return dp[i][j];
    }
     
    dp[i][j] = Number.MAX_VALUE;
    for(let k = i; k < j; k++) 
    {
        dp[i][j] = Math.min(
            dp[i][j], matrixChainMemoised(p, i, k) +
                      matrixChainMemoised(p, k + 1, j) +
                      p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
 
function MatrixChainOrder(p, n)
{
    let i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}
 
// Driver code
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;
 
for(var i = 0; i < dp.length; i++)
{
    for(var j = 0; j < dp.length; j++)
    {
        dp[i][j] = -1;
    }
}
 
document.write("Minimum number of multiplications is " +
               MatrixChainOrder(arr, n));
                
// This code is contributed by target_2
 

输出

Minimum number of multiplications is 18

使用制表 –

C++

// See the Cormen book for details of the
// following algorithm
#include 
using namespace std;
 
// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1..n
int MatrixChainOrder(int p[], int n)
{
 
    /* For simplicity of the program, one
    extra row and one extra column are
    allocated in m[][]. 0th row and 0th
    column of m[][] are not used */
    int m[n][n];
 
    int i, j, k, L, q;
 
    /* m[i, j] = Minimum number of scalar
    multiplications needed to compute the
    matrix A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i] */
 
    // cost is zero when multiplying
    // one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, size);
 
    getchar();
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// See the Cormen book for details of the following
// algorithm
#include 
#include 
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int MatrixChainOrder(int p[], int n)
{
 
    /* For simplicity of the program,
       one extra row and one
       extra column are allocated in m[][]. 
       0th row and 0th
       column of m[][] are not used */
    int m[n][n];
 
    int i, j, k, L, q;
 
    /* m[i, j] = Minimum number of
       scalar multiplications
       needed to compute the matrix
       A[i]A[i+1]...A[j] =
       A[i..j] where dimension of A[i]
       is p[i-1] x p[i] */
 
    // cost is zero when multiplying one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++) {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver  code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, size));
 
    getchar();
    return 0;
}

Java

// Dynamic Programming Java implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm
class MatrixChainMultiplication
{
 
    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    static int MatrixChainOrder(int p[], int n)
    {
        /* For simplicity of the
        program, one extra row and
        one extra column are allocated in m[][].  0th row
        and 0th column of m[][] are not used */
        int m[][] = new int[n][n];
 
        int i, j, k, L, q;
 
        /* m[i, j] = Minimum number of scalar
        multiplications needed to compute the matrix
        A[i]A[i+1]...A[j] = A[i..j] where
        dimension of A[i] is p[i-1] x p[i] */
 
        // cost is zero when multiplying one matrix.
        for (i = 1; i < n; i++)
            m[i][i] = 0;
 
        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i][j] = Integer.MAX_VALUE;
                for (k = i; k <= j - 1; k++)
                {
                    // q = cost/scalar multiplications
                    q = m[i][k] + m[k + 1][j]
                        + p[i - 1] * p[k] * p[j];
                    if (q < m[i][j])
                        m[i][j] = q;
                }
            }
        }
 
        return m[1][n - 1];
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4 };
        int size = arr.length;
 
        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, size));
    }
}
/* This code is contributed by Rajat Mishra*/

Python

# Dynamic Programming Python implementation of Matrix
# Chain Multiplication. See the Cormen book for details
# of the following algorithm
import sys
 
# Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
 
 
def MatrixChainOrder(p, n):
    # For simplicity of the program,
    # one extra row and one
    # extra column are allocated in m[][]. 
    # 0th row and 0th
    # column of m[][] are not used
    m = [[0 for x in range(n)] for x in range(n)]
 
    # m[i, j] = Minimum number of scalar
    # multiplications needed
    # to compute the matrix A[i]A[i + 1]...A[j] =
    # A[i..j] where
    # dimension of A[i] is p[i-1] x p[i]
 
    # cost is zero when multiplying one matrix.
    for i in range(1, n):
        m[i][i] = 0
 
    # L is chain length.
    for L in range(2, n):
        for i in range(1, n-L + 1):
            j = i + L-1
            m[i][j] = sys.maxint
            for k in range(i, j):
 
                # q = cost / scalar multiplications
                q = m[i][k] + m[k + 1][j] + p[i-1]*p[k]*p[j]
                if q < m[i][j]:
                    m[i][j] = q
 
    return m[1][n-1]
 
 
# Driver code
arr = [1, 2, 3, 4]
size = len(arr)
 
print("Minimum number of multiplications is " +
      str(MatrixChainOrder(arr, size)))
# This Code is contributed by Bhavya Jain

C＃

// Dynamic Programming C# implementation of
// Matrix Chain Multiplication.
// See the Cormen book for details of the
// following algorithm
using System;
 
class GFG
{
 
    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int n)
    {
 
        /* For simplicity of the program, one
        extra row and one extra column are
        allocated in m[][]. 0th row and 0th
        column of m[][] are not used */
        int[, ] m = new int[n, n];
 
        int i, j, k, L, q;
 
        /* m[i, j] = Minimum number of scalar
        multiplications needed
        to compute the matrix A[i]A[i+1]...A[j]
        = A[i..j] where dimension of A[i] is
        p[i-1] x p[i] */
 
        // cost is zero when multiplying
        // one matrix.
        for (i = 1; i < n; i++)
            m[i, i] = 0;
 
        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i, j] = int.MaxValue;
                for (k = i; k <= j - 1; k++)
                {
                    // q = cost/scalar multiplications
                    q = m[i, k] + m[k + 1, j]
                        + p[i - 1] * p[k] * p[j];
                    if (q < m[i, j])
                        m[i, j] = q;
                }
            }
        }
 
        return m[1, n - 1];
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4 };
        int size = arr.Length;
 
        Console.Write("Minimum number of "
                      + "multiplications is "
                      + MatrixChainOrder(arr, size));
    }
}
 
// This code is contributed by Sam007

Javascript

 
// Dynamic Programming javascript implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm
 
 
// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(p , n)
{
    /* For simplicity of the
    program, one extra row and
    one extra column are allocated in m.  0th row
    and 0th column of m are not used */
    var m = Array(n).fill(0).map(x => Array(n).fill(0));
 
    var i, j, k, L, q;
 
    /* m[i, j] = Minimum number of scalar
    multiplications needed to compute the matrix
    A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i] */
 
    // cost is zero when multiplying one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;
 
    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            if (j == n)
                continue;
            m[i][j] = Number.MAX_VALUE;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }
 
    return m[1][n - 1];
}
 
// Driver code
var arr = [ 1, 2, 3, 4 ];
var size = arr.length;
 
document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, size));
 
// This code contributed by Princi Singh
 

输出

Minimum number of multiplications is 18

时间复杂度： O(n ³ )
辅助空间： O(n ² )

应用：
带有 * 和 + 的表达式的最小值和最大值

参考：
http://en.wikipedia.org/wiki/Matrix_chain_multiplication
http://www.personal.kent.edu/~rmuhamma/Algorithms/MyAlgorithms/Dynamic/chainMatrixMult.htm

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