动态规划：矩阵连乘问题

下面仅仅是对此问题的一个代码实现，详细理论部分请參见王晓东《算法设计与分析》第2版3.1节 矩阵连乘问题。

#include <iostream>
#include <iomanip>


using namespace std;

#define MAX_COUNT			20

//矩阵属性
struct tagMatrixAttribute
{
	int row;
	int col;
};

//矩阵连乘加括号求解
void MatrixChain( tagMatrixAttribute mMatrix[], int nCount,
				 int m[][MAX_COUNT], int s[][MAX_COUNT] )
{
	for ( int i = 0; i < nCount; ++i ) m[i][i] = 0;

	for ( int r = 1; r < nCount; ++r )
	{
		for ( int i = 0; i < nCount - r; ++i )
		{
			int j = i + r;
			//从k处断开，i <= k < j
			int k = i;
			m[i][j] = m[i][k] + m[k+1][j] 
			+ mMatrix[i].row * mMatrix[k].col * mMatrix[j].col;
			s[i][j] = k;
			int nTemp = 0;
			for( k = i + 1; k < j; ++k )
			{
				nTemp = m[i][k] + m[k+1][j] 
				+ mMatrix[i].row * mMatrix[k].col * mMatrix[j].col;
				if ( nTemp < m[i][j] )
				{
					m[i][j] = nTemp;
					s[i][j] = k;
				}
			}
		}
	}
}

//构造结果 
void TraceBack( int s[][MAX_COUNT], int i, int j )
{
	if ( i == j ) 
	{
		cout << "A" << i+1;
		return;
	}

	cout << "(";
	TraceBack( s, i, s[i][j] );
	TraceBack( s, s[i][j]+1, j );
	cout << ")";
}


void PrintArray( int nArray[][MAX_COUNT], int nCount )
{
	cout << left;
	for( int i = 0; i < nCount; ++i )
	{
		for ( int j = 0; j < nCount; ++j )
		{
			cout << setw(7) << nArray[i][j] << " ";
		}
		cout << endl;
	}
	cout << right;
}


int main()
{
	tagMatrixAttribute mMatrixAttrArray[] = {
		30, 35,
		35, 15,
		15, 5,
		5, 10,
		10, 20,
		20, 25
	};
// 	tagMatrixAttribute mMatrixAttrArray[] = {
// 		10, 100,
// 		100, 5,
// 		5, 50
// 	};
	int nCount = _countof( mMatrixAttrArray );

	int m[MAX_COUNT][MAX_COUNT];
	int s[MAX_COUNT][MAX_COUNT];
	memset( m, 0, sizeof(m) );
	memset( s, 0, sizeof(s) );

	MatrixChain( mMatrixAttrArray, nCount, m, s );

	PrintArray( m, nCount );
	cout << endl;

	PrintArray( s, nCount );
	cout << endl;

	//构造结果
	TraceBack( s, 0, nCount-1 );
	cout << endl;

	return 0;
}

作者：山丘儿
转载请标明出处。谢谢。

原文地址：http://blog.csdn.net/s634772208/article/details/46683087