算法导论学习——分治矩阵乘法

头文件 结构的定义

stdafx.h

// stdafx.h : 标准系统包含文件的包含文件，
// 或是经常使用但不常更改的
// 特定于项目的包含文件
//

#pragma once

#include "targetver.h"

#include <stdio.h>
#include <tchar.h>



// TODO:  在此处引用程序需要的其他头文件
#include <iostream>

using namespace std;


#define Maxelem 10

//求两个数的最大值
int inline max(int a, int b){
	return a >= b ? a : b;
}
//求三个数的最大值
int inline max(int a, int b, int c){
	return max(max(a, b), c);
}

//求min(2^x),ST. 2^x>=number。
int inline L2n(int number){
	if ((number & number - 1) == 0)
	{
		return number;
	}
	else {
		return pow(2, (int)log2(number) + 1);
	}
}

//定义矩阵的结构
class Matrix{

	
public:
	int data[Maxelem][Maxelem];
	int M, N;
	//以数组复制的方式构造矩阵对象，其中起始位置为0.
	Matrix(int array[Maxelem][Maxelem], int m, int n){
		M = m;
		N = n;
		for (int i = 0; i < m; i++){
			for (int j = 0; j < n; j++){
				this->data[i][j] = array[i][j];
			}
		}
	}
	
	//以数组复制的方式构造矩阵对象，以strat1,end1,strat2,end2为区间
	Matrix(int array[Maxelem][Maxelem], int start1, int end1, int start2, int end2){
		M = end1 - start1 + 1;
		N = end2 - start2 + 1;
		int p = 0, q = 0;
		for (int i = 0; i < M; i++){
			for (int j = 0; j < N; j++){
				data[i][j] = array[start1 + i][start2 +j];
			}
			
		}
			
	}
	//仅构造矩阵，不填充数据
	Matrix(int m, int n) :M(m), N(n){}


	/*
	 打印输出
	*/
	void print() const {
		for (int i = 0; i <M; i++){
			for (int j = 0; j < N; j++){
				cout << data[i][j] << " ";
			} 
			cout << endl;

		}
	}
	//为矩阵补充0，使其成为标准方阵
		void fill(){
		
		if (!(M == N && ((M & M - 1) == 0))){
			int n = L2n(max(M, N));
			
			for (int i = 0; i < n; i++){
				for (int j = 0; j < n; j++){
					data[i][j] = (i < M&&j < N ? data[i][j] : 0);
				}
			}
			M = n;
			N = n;
			
		}
	}


	/*重载二维运算符[][]*/
	int * const operator[](const int i)
	{
		return data[i];
	}

	Matrix friend operator +(Matrix m1, Matrix m2){

	
		Matrix op = Matrix(m1.M, m1.N);
		for (int i = 0; i < m1.M; i++)
		{
			for (int j = 0; j < m1.M; j++){
				op[i][j] = m1[i][j] + m2[i][j];
			}
		}
		return op;
	}
	Matrix friend operator -(Matrix m1, Matrix m2){


		Matrix op = Matrix(m1.M, m1.N);
		for (int i = 0; i < m1.M; i++)
		{
			for (int j = 0; j < m1.M; j++){
				op[i][j] = m1[i][j] - m2[i][j];
			}
		}
		return op;
	}
	/*
	将计算过程中补充的0清除。计算完毕后才能用的方法，不加也能得到结果，不过行列数不对。
	*/
	void clean(int m, int n){
		M = m;
		N = n;
	}
};

　　算法的实现：

// strassenAlgorithm.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"




//该方法只能相乘最简单的2*2矩阵。
Matrix mutilSimple(Matrix A, Matrix B){
	int a = A[0][0],b=A[0][1],c=A[1][0],d=A[1][1];
	int e = B[0][0], f = B[0][1], g = B[1][0], h = B[1][1];

	int p1 = a*(f - h);
	int p2 = (a + b)*h;
	int p3 = (c + d)*e;
	int p4 = d*(g - e);
	int p5 = (a + d)*(e + h);
	int p6 = (b - d)*(g + h);
	int p7 = (a - c)*(e + f);
	
	Matrix returnValue = Matrix(2, 2);
	returnValue[0][0] = p5 + p4 - p2 + p6;
	returnValue[0][1] = p1 + p2;
	returnValue[1][0] = p3 + p4;
	returnValue[1][1] = p1 + p5 - p3 - p7;

	return returnValue;
}

//矩阵乘法 必须用了fill方法才能相乘
Matrix mutilMerge(Matrix A,Matrix B){
	if (A.M == 2 && B.M == 2){
		return mutilSimple(A, B);
	}
	int k = A.M;
	Matrix 
		a = Matrix(A.data, 0, k / 2 - 1, 0, k / 2 - 1),
		b = Matrix(A.data, 0, k / 2 - 1, k / 2, k - 1),
		c = Matrix(A.data, k / 2, k - 1, 0, k / 2 - 1),
		d = Matrix(A.data, k / 2, k - 1, k / 2, k - 1),
	
		e = Matrix(B.data, 0, k / 2 - 1, 0, k / 2 - 1),
		f = Matrix(B.data, 0, k / 2 - 1, k / 2, k - 1),
		g = Matrix(B.data, k / 2, k - 1, 0, k / 2 - 1),
		h = Matrix(B.data, k / 2, k - 1, k / 2, k - 1),
		op = Matrix(k, k),
	
		p1 = mutilMerge(a, f - h),
		p2 = mutilMerge(a + b, h),
		p3 = mutilMerge(c + d, e),
		p4 = mutilMerge(d, g - e),
		p5 = mutilMerge(a + d, e + h),
		p6 = mutilMerge(b - d, g + h),
		p7 = mutilMerge(a - c, e + f),
	
		
		op1=p5+p4-p2+p6,
		op2=p1+p2,
		op3=p3+p4,
		op4 = p1 + p5 - p3 - p7; 



	int x1 = 0, y1 = 0, x2 = 0, y2 = 0, x3 = 0,y3=0,x4=0, y4 = 0; //4个变量的游标
	int u = 0, v = 0;
	for (int i = 0; i < k; i++)
	{
		for (int j = 0; j < k; j++){
		
			if (i >= 0 && i <= k / 2 - 1 && j >= 0 && j <= k / 2 - 1){
				op[i][j] = op1[x1][y1];
				y1++;
				if (y1 == op1.M) { y1 = 0; x1++; }
			}
			if (i >= 0 && i <= k / 2 - 1 && j >= k / 2 && j <= k - 1){
				op[i][j] = op2[x2][y2];
				y2++;
				if (y2 == op2.M) { y2 = 0; x2++; }
			}
			if (i >= k/2 && i <= k - 1 && j >= 0 && j <= k / 2 - 1){

				op[i][j] = op3[x3][y3];
				y3++;
				if (y3 == op3.M) { y3 = 0; x3++; }
			}
			
			if (i >= k / 2 && i <= k - 1 && j >= k/2 && j <= k - 1){
				op[i][j] = op4[x4][y4];
				y4++;
				if (y4 == op4.M) { y4 = 0; x4++; }
			}

		}
	}


	
	return op;
	
}
Matrix zeroclear( Matrix result,int M,int N){
	result.M = M;
	result.N = N;
	return result;
}
int _tmain(int argc, _TCHAR* argv[])
{

	int matrixA[Maxelem][Maxelem] = {
		{ 10, 3, 3 ,7,4},
		{ 5, 3, 8,2,1 },
		{ -2, 3, 7, 5, 2 }, 
		{1,10,-2,1,8},
		{3,3,3,3,3}
	};

	int matrixB[Maxelem][Maxelem] = {
		{ -4, 6, 1,2,1 },
		{ 9, 10, 8,0,3 },
		{ 2, 3, -7 ,-1,-1},
		{ 1, -6, 2, 1, 7 }, 
		{1,2,3,4,5}
	};

	Matrix ma = Matrix(matrixA,5,5);
	Matrix mb = Matrix(matrixB,5,5);

	
	cout << "A="<<endl;
	
	ma.print();
	cout << "B="<<endl;
	mb.print();

	ma.fill();
	mb.fill();
	Matrix result = mutilMerge(ma, mb);
	
	cout << "A×B=" << endl;
	result.clean(5, 5);
	result.print();
	cout << "B×A=" << endl;
	result = mutilMerge(mb, ma);
	result.clean(5, 5);
	result.print();
	system("pause");

	return 0;
}