n阶矩阵一般乘法-《算法导论》学习笔记五

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A、B两个矩阵均是nxn的矩阵，则两个矩阵的乘法：
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一般的矩阵乘法代码：

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <unistd.h>#include <time.h>class SquareMatrix {public:    SquareMatrix(){}    SquareMatrix( int row, int col ):        mRow(row), mCol(col)    {        Init();    }    ~SquareMatrix()    {        for ( int i = 0; i < mRow; i++ ) {            delete[] mElement[i];            //printf("delete %d\n", i);        }        delete[] mElement;    }    void Init()    {        mElement = new int*[mRow];        for ( int i = 0; i < mCol; i++ ) {            mElement[i] = new int[i];        }    }    void SetElement( int lowV, int upV )    {        int size = upV - lowV;        for ( int i = 0; i < mRow; i++ ) {            for ( int j = 0; j < mCol; j++ ) {                mElement[i][j] = rand() % size + lowV;            }        }    }    void PrintElement()    {        printf("=========================================\n");        for ( int i = 0; i < mRow; i++ ) {            for ( int j = 0; j < mCol; j++ ) {                printf("%d ", mElement[i][j]);            }            printf("\n");        }        printf("=========================================\n");    }public:    int mRow;    int mCol;    int *(*mElement);};void SquareMatrixMultiply(     SquareMatrix &a,     SquareMatrix &b,     SquareMatrix &c ){    int n = a.mRow;    c.mRow = c.mCol = n;    c.Init();    for ( int i = 0; i < n; i++ ) {        for ( int j = 0; j < n; j++ ) {            c.mElement[i][j] = 0;            for ( int k = 0; k < n; k++ ) {                c.mElement[i][j] += a.mElement[i][k] * b.mElement[j][k];            }        }    }}int main( int argc, char **argv ) {    if ( argc != 2 ) {        printf("Usage:./binaryfile num\n");        exit( 0 );    }    int n = atoi( argv[1] );    SquareMatrix smA( n, n ), smB( n, n ), smC;    smA.SetElement( 1, 10 );    smA.PrintElement();    smB.SetElement( 1, 10 );    smB.PrintElement();    SquareMatrixMultiply( smA, smB, smC );    smC.PrintElement();    printf("init square matrix finished\n");    return 0;}

算法复杂度为O(n^3)，而Stranssen算法通过分治法将大矩阵切分为小矩阵进行计算，算法复杂度可以降低为O(n^2.81)，但是尝试写下代码，发现切割子矩阵时有点复杂，普通的切分会创建子矩阵并复制值，而用下标进行计算又比较复杂，下次有空再尝试写吧。

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