Geeks面试题：Matrix Chain Multiplication

Matrix Chain Multiplication

Given a sequence of matrices, find the most efficient way to multiply these matrices together.The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications.

We have many options to multiply a chain of matrices because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B, C, and D, we would have:

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

However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. For example, suppose A is a 10 × 30 matrix, B is a 30 × 5 matrix, and C is a 5 × 60 matrix. Then,

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

Clearly the first method is the more efficient.

Given an array p[] which represents the chain of matrices such that the ith matrix Ai is of dimension p[i-1] x p[i]. We need to write a function MatrixChainOrder() that should return the minimum number of multiplications needed to multiply the chain.

  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

http://www.geeksforgeeks.org/dynamic-programming-set-8-matrix-chain-multiplication/

斜对角线的遍历真的好麻烦，需要非常小心跟踪下标的走势。

递归法和动态规划法程序：

// 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 multiplcations 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;
}

//第一个p的数值代表最左边的一个行，其他p的数组代表列，p的元素比矩阵数多一的。即有n-1个矩阵，p要存储n个元素
//i，j为两边定点，然后以d为段，段越大，活动空间越小，以k为i到j之间，斜角线遍历
//关键要定位好i和j，k就好办了。记得要由段来计算的，因为矩阵根据段d来活动。
//思路要清晰，才能跟踪好下标。
int matrixChainDP(int p[], int n)
{
	vector<vector<int> > m(n-1, vector<int>(n-1));

	for (int d = 2; d <= n-1; d++)
	{
		for (int i = 0; i < n-d; i++)
		{
			int j = i+d-1;
			m[i][j] = INT_MAX;
			for (int k = i; k < j; k++)
			{
				int t = m[i][k] + m[k+1][j] + p[i]*p[k+1]*p[j+1];
				m[i][j] = min(t, m[i][j]);
			}
		} 
	}
	return m[0][n-2];
}


int main()
{
	int arr[] = {1, 2, 3, 4};
	int size = sizeof(arr)/sizeof(arr[0]);

	printf("Minimum number of multiplications is %d ",
		MatrixChainOrder(arr, 1, size-1));

	cout<<endl<<matrixChainDP(arr, size)<<endl;

	system("pause");
	return 0;
}