算法导论实验——递归与分治

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实验内容:
1.复习递归与分治的基本知识,了解分治的基本思想;
2.利用分治思想设计并解决常见问题,如矩阵相乘,循环赛日程表等

操作环境:
操作系统:Win XP/Win 7/Win8 操作系统
编程语言:C 语言
开发工具:e.g.,Microsoft Visual C++ 6.0

1 递归与分治算法设计与实现
问题描述:将正整数 n 表示成一系列正整数之和:n=n1+n2+…+nk,其中 n1≥n2 ≥…≥nk≥1,k≥1。正整数 n 的这种表示称为正整数 n 的划分。求正整数 n 的不同划分个数。列出算法设计过程,写出实现代码。

实现代码:

#include 
#include 
using namespace std;

//m为小于Number的正整数,NumDivide函数计算正整数m的划分中加数小于或等于m的所有划分
int NumDivide(int Number, int m){
	if (Number < 1 || m < 1) return 0;
	else if (Number == 1 || m == 1) return 1;
	else if (Number < m) return NumDivide(Number, Number);
	else if (Number == m) return NumDivide(Number, Number - 1) + 1;
	else return NumDivide(Number, m - 1) + NumDivide(Number - m, m);
}

int main(void){
	int Number,m,DivideTimes;
	cout << "输入数字Number:" << endl;
	cin >> Number;
	DivideTimes = NumDivide(Number, Number - 1);
	cout << "一共划分" << DivideTimes << "次" << endl;
	return 0;
}

算法导论实验——递归与分治_第1张图片

2.Strassen 矩阵乘法的 C 语言实现
问题描述: 求解两个矩阵 A 和 B 的乘积,C=AB,可将矩阵分块
在这里插入图片描述
根据矩阵乘法,可知矩阵 C 的计算公式:
在这里插入图片描述
以上过程有 8 次乘法运算,4 次加法运算,矩阵乘法的复杂度主要就是体现在相乘上,而多一两次的加法并不会让复杂度上升太多。故此,是否可以让矩阵乘法的运算过程中乘法的运算次数减少?

现定义如下 7 个新矩阵:
算法导论实验——递归与分治_第2张图片
那么,矩阵 C 则可以通过以上 7 个矩阵计算得到
在这里插入图片描述
主要代码实现:

#include 
#include 
#include 
#include 
using namespace std;

template//函数模板声明
class Strassen_class {
public:
	void ADD(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize);
	void SUB(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize);
	void MUL(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize);//朴素算法实现
	void FillMatrix(T** MatrixA, T** MatrixB, int length);//A,B矩阵赋值
	void PrintMatrix(T **MatrixA, int MatrixSize);//打印矩阵
	void Strassen(int N, T **MatrixA, T **MatrixB, T **MatrixC);//Strassen算法实现
};
template
void Strassen_class::ADD(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize)
{
	for (int i = 0; i < MatrixSize; i++)
	{
		for (int j = 0; j < MatrixSize; j++)
		{
			MatrixResult[i][j] = MatrixA[i][j] + MatrixB[i][j];
		}
	}
}
template
void Strassen_class::SUB(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize)
{
	for (int i = 0; i < MatrixSize; i++)
	{
		for (int j = 0; j < MatrixSize; j++)
		{
			MatrixResult[i][j] = MatrixA[i][j] - MatrixB[i][j];
		}
	}
}
template
void Strassen_class::MUL(T** MatrixA, T** MatrixB, T** MatrixResult, int MatrixSize)
{
	for (int i = 0; i
void Strassen_class::Strassen(int N, T **MatrixA, T **MatrixB, T **MatrixC)
{
	int HalfSize = N / 2;
	int newSize = N / 2;

	if (N <= 32)    //分治门槛,小于这个值时不再进行递归计算,而是采用常规矩阵计算方法
	{
		MUL(MatrixA, MatrixB, MatrixC, N);
	}
	else
	{
		T** A11;
		T** A12;
		T** A21;
		T** A22;

		T** B11;
		T** B12;
		T** B21;
		T** B22;

		T** C11;
		T** C12;
		T** C21;
		T** C22;

		T** M1;
		T** M2;
		T** M3;
		T** M4;
		T** M5;
		T** M6;
		T** M7;
		T** AResult;
		T** BResult;

		//making a 1 diminsional pointer based array.
		A11 = new T *[newSize];
		A12 = new T *[newSize];
		A21 = new T *[newSize];
		A22 = new T *[newSize];

		B11 = new T *[newSize];
		B12 = new T *[newSize];
		B21 = new T *[newSize];
		B22 = new T *[newSize];

		C11 = new T *[newSize];
		C12 = new T *[newSize];
		C21 = new T *[newSize];
		C22 = new T *[newSize];

		M1 = new T *[newSize];
		M2 = new T *[newSize];
		M3 = new T *[newSize];
		M4 = new T *[newSize];
		M5 = new T *[newSize];
		M6 = new T *[newSize];
		M7 = new T *[newSize];

		AResult = new T *[newSize];
		BResult = new T *[newSize];

		int newLength = newSize;

		//making that 1 diminsional pointer based array , a 2D pointer based array
		for (int i = 0; i < newSize; i++)
		{
			A11[i] = new T[newLength];
			A12[i] = new T[newLength];
			A21[i] = new T[newLength];
			A22[i] = new T[newLength];

			B11[i] = new T[newLength];
			B12[i] = new T[newLength];
			B21[i] = new T[newLength];
			B22[i] = new T[newLength];

			C11[i] = new T[newLength];
			C12[i] = new T[newLength];
			C21[i] = new T[newLength];
			C22[i] = new T[newLength];

			M1[i] = new T[newLength];
			M2[i] = new T[newLength];
			M3[i] = new T[newLength];
			M4[i] = new T[newLength];
			M5[i] = new T[newLength];
			M6[i] = new T[newLength];
			M7[i] = new T[newLength];

			AResult[i] = new T[newLength];
			BResult[i] = new T[newLength];


		}
		//splitting input Matrixes, into 4 submatrices each.
		for (int i = 0; i < N / 2; i++)
		{
			for (int j = 0; j < N / 2; j++)
			{
				A11[i][j] = MatrixA[i][j];
				A12[i][j] = MatrixA[i][j + N / 2];
				A21[i][j] = MatrixA[i + N / 2][j];
				A22[i][j] = MatrixA[i + N / 2][j + N / 2];

				B11[i][j] = MatrixB[i][j];
				B12[i][j] = MatrixB[i][j + N / 2];
				B21[i][j] = MatrixB[i + N / 2][j];
				B22[i][j] = MatrixB[i + N / 2][j + N / 2];

			}
		}

		//here we calculate M1..M7 matrices .
		//M1[][]
		ADD(A11, A22, AResult, HalfSize);
		ADD(B11, B22, BResult, HalfSize);                //p5=(a+d)*(e+h)
		Strassen(HalfSize, AResult, BResult, M1); //now that we need to multiply this , we use the strassen itself .


		//M2[][]
		ADD(A21, A22, AResult, HalfSize);              //M2=(A21+A22)B11   p3=(c+d)*e
		Strassen(HalfSize, AResult, B11, M2);       //Mul(AResult,B11,M2);

		//M3[][]
		SUB(B12, B22, BResult, HalfSize);              //M3=A11(B12-B22)   p1=a*(f-h)
		Strassen(HalfSize, A11, BResult, M3);       //Mul(A11,BResult,M3);

		//M4[][]
		SUB(B21, B11, BResult, HalfSize);           //M4=A22(B21-B11)    p4=d*(g-e)
		Strassen(HalfSize, A22, BResult, M4);       //Mul(A22,BResult,M4);

		//M5[][]
		ADD(A11, A12, AResult, HalfSize);           //M5=(A11+A12)B22   p2=(a+b)*h
		Strassen(HalfSize, AResult, B22, M5);       //Mul(AResult,B22,M5);


		//M6[][]
		SUB(A21, A11, AResult, HalfSize);
		ADD(B11, B12, BResult, HalfSize);             //M6=(A21-A11)(B11+B12)   p7=(c-a)(e+f)
		Strassen(HalfSize, AResult, BResult, M6);    //Mul(AResult,BResult,M6);

		//M7[][]
		SUB(A12, A22, AResult, HalfSize);
		ADD(B21, B22, BResult, HalfSize);             //M7=(A12-A22)(B21+B22)    p6=(b-d)*(g+h)
		Strassen(HalfSize, AResult, BResult, M7);     //Mul(AResult,BResult,M7);

		//C11 = M1 + M4 - M5 + M7;
		ADD(M1, M4, AResult, HalfSize);
		SUB(M7, M5, BResult, HalfSize);
		ADD(AResult, BResult, C11, HalfSize);

		//C12 = M3 + M5;
		ADD(M3, M5, C12, HalfSize);

		//C21 = M2 + M4;
		ADD(M2, M4, C21, HalfSize);

		//C22 = M1 + M3 - M2 + M6;
		ADD(M1, M3, AResult, HalfSize);
		SUB(M6, M2, BResult, HalfSize);
		ADD(AResult, BResult, C22, HalfSize);

		//at this point , we have calculated the c11..c22 matrices, and now we are going to
		//put them together and make a unit matrix which would describe our resulting Matrix.
		//组合小矩阵到一个大矩阵
		for (int i = 0; i < N / 2; i++)
		{
			for (int j = 0; j < N / 2; j++)
			{
				MatrixC[i][j] = C11[i][j];
				MatrixC[i][j + N / 2] = C12[i][j];
				MatrixC[i + N / 2][j] = C21[i][j];
				MatrixC[i + N / 2][j + N / 2] = C22[i][j];
			}
		}

		// 释放矩阵内存空间
		for (int i = 0; i < newLength; i++)
		{
			delete[] A11[i]; delete[] A12[i]; delete[] A21[i];
			delete[] A22[i];

			delete[] B11[i]; delete[] B12[i]; delete[] B21[i];
			delete[] B22[i];
			delete[] C11[i]; delete[] C12[i]; delete[] C21[i];
			delete[] C22[i];
			delete[] M1[i]; delete[] M2[i]; delete[] M3[i]; delete[] M4[i];
			delete[] M5[i]; delete[] M6[i]; delete[] M7[i];
			delete[] AResult[i]; delete[] BResult[i];
		}
		delete[] A11; delete[] A12; delete[] A21; delete[] A22;
		delete[] B11; delete[] B12; delete[] B21; delete[] B22;
		delete[] C11; delete[] C12; delete[] C21; delete[] C22;
		delete[] M1; delete[] M2; delete[] M3; delete[] M4; delete[] M5;
		delete[] M6; delete[] M7;
		delete[] AResult;
		delete[] BResult;

	}//end of else

}

template
void Strassen_class::FillMatrix(T** MatrixA, T** MatrixB, int length)
{
	cout << "输入A矩阵的值" << endl;
	for (int row = 0; row> MatrixA[row][column];

			//MatrixB[row][column] = (MatrixA[row][column] = rand() % 5);
			//matrix2[row][column] = rand() % 2;//ba hazfe in khat 50% afzayeshe soorat khahim dasht
		}

	}

	cout << "输入矩阵B的值" << endl;
	for (int row = 0; row> MatrixB[row][column];

			//MatrixB[row][column] = (MatrixA[row][column] = rand() % 5);
			//matrix2[row][column] = rand() % 2;//ba hazfe in khat 50% afzayeshe soorat khahim dasht
		}

	}
}
template
void Strassen_class::PrintMatrix(T **MatrixA, int MatrixSize)
{
	cout << endl;
	for (int row = 0; row stra;//定义Strassen_class类对象
	int MatrixSize = 0;

	int** MatrixA;    //存放矩阵A
	int** MatrixB;    //存放矩阵B
	int** MatrixC;    //存放结果矩阵

	cout << "\n请输入矩阵大小(必须是2的幂指数值): ";
	cin >> MatrixSize;

	int N = MatrixSize;//for readiblity.

	//申请内存
	MatrixA = new int *[MatrixSize];
	MatrixB = new int *[MatrixSize];
	MatrixC = new int *[MatrixSize];

	for (int i = 0; i < MatrixSize; i++)
	{
		MatrixA[i] = new int[MatrixSize];
		MatrixB[i] = new int[MatrixSize];
		MatrixC[i] = new int[MatrixSize];
	}

	stra.FillMatrix(MatrixA, MatrixB, MatrixSize);  //矩阵赋值

	//stra.MUL(MatrixA, MatrixB, MatrixC, MatrixSize);//朴素矩阵相乘算法 T(n) = O(n^3)

	stra.Strassen(N, MatrixA, MatrixB, MatrixC); //strassen矩阵相乘算法
	cout << "\n矩阵运算结果... \n";
	stra.PrintMatrix(MatrixC, MatrixSize);//打印矩阵

	system("Pause");

	return 0;
}

运行结果:
算法导论实验——递归与分治_第3张图片

结果分析与讨论:
当矩阵维数比较小(小于8时)使用简易算法,即直接进行矩阵相乘与相加所用时间比较短,效率高;当矩阵维数比较大(大于8)使用Strassen算法更加高效,时间复杂度低,运行速度快。

3.循环赛日程表问题问题描述:要求设计满足以下要求的比赛日程表

(1)每个选手必须与其他 n-1 个选手各赛一次;
(2)每个选手一天只能赛一次;
(3)循环赛一共进行 n-1 天。

算法导论实验——递归与分治_第4张图片
实现提示:
按分治策略,将所有的选手分为两半,n 个选手的比赛日程表就可以通过为 n/2 个选手设计的比赛日程表来决定。递归地用对选手进行分割,直到只剩下 2 个选手时,比赛日程表的制定就变得很简单。这时只要让这 2 个选手进行比赛就可以了。

程序源代码:

xunhuan.h
#include
#include
using namespace std;

void GameTable(vector > &vec);
void start();

xunhuan.cpp
#include "xunhuan.h"

void GameTable(vector > &vec){
	if (vec.size() == 0){
		return;
	}
	size_t s = vec.size();
	int k = 0;
	while (s = s >> 1){
		//s = s >> 1;
		k++;
	}
	//初始化
	vec[0][0] = 1;
	vec[0][1] = 2;
	vec[1][0] = 2;
	vec[1][1] = 1;

	for (int i = 2; i <= k; i++){
		int length = 0x1 << i;
		int half = length >> 1;
		//左下角的子表中项为左上角子表对应项加half=2^(i-1)
		for (int row = 0; row < half; row++){
			for (int col = 0; col < half; col++){
				vec[row + half][col] = vec[row][col] + half;
			}
		}
		//右上角的子表等于左下角子表
		for (int row = 0; row < half; row++){
			for (int col = 0; col < half; col++){
				vec[row][col + half] = vec[row + half][col];
			}
		}
		//右下角的子表等于左上角子表
		for (int row = 0; row < half; row++){
			for (int col = 0; col < half; col++){
				vec[row + half][col + half] = vec[row][col];
			}
		}
	}
}
void start(){
	cout << "共有2^k个选手参加比赛,输入k(k>0):" << endl;
	int k;
	do{
		cin >> k;
	} while (k < 0 || k > 31);

	int s = 0x1 << k;
	vector > vec(s, vector(s, 0));

	GameTable(vec);

	for (size_t i = 0; i < vec.size(); i++){
		for (size_t j = 0; j < vec[i].size(); j++){
			cout << vec[i][j] << " ";
		}
		cout << endl;
	}
}

main.cpp
#include "xunhuan.h"

int main(void){
	start();
	system("Pause");
	return 0;
}

运行结果

算法导论实验——递归与分治_第5张图片

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