大家如果对算法导论以及其他信息感兴趣，可以加入学习交流群一起交流 948097478

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 <stdio.h>
#include <iostream>
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;
}

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

根据矩阵乘法，可知矩阵 C 的计算公式：

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：

那么,矩阵 C 则可以通过以上 7 个矩阵计算得到

主要代码实现：

#include <tchar.h>
#include <iostream>
#include <ctime>
#include <Windows.h>
using namespace std;template<typename T>//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<typename T>
void Strassen_class<T>::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<typename T>
void Strassen_class<T>::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<typename T>
void Strassen_class<T>::MUL(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] = 0;for (int k = 0; k<MatrixSize; k++){MatrixResult[i][j] = MatrixResult[i][j] + MatrixA[i][k] * MatrixB[k][j];}}}
}template<typename T>
void Strassen_class<T>::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 arrayfor (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)*eStrassen(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)*hStrassen(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<typename T>
void Strassen_class<T>::FillMatrix(T** MatrixA, T** MatrixB, int length)
{cout << "输入A矩阵的值" << endl;for (int row = 0; row<length; row++){for (int column = 0; column<length; column++){cin >> 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<length; row++){for (int column = 0; column<length; column++){cin >> 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<typename T>
void Strassen_class<T>::PrintMatrix(T **MatrixA, int MatrixSize)
{cout << endl;for (int row = 0; row<MatrixSize; row++){for (int column = 0; column<MatrixSize; column++){cout << MatrixA[row][column] << "\t";if ((column + 1) % ((MatrixSize)) == 0)cout << endl;}}cout << endl;
}int _tmain(int argc, _TCHAR* argv[])
{Strassen_class<int> stra;//定义Strassen_class类对象int MatrixSize = 0;int** MatrixA;    //存放矩阵Aint** MatrixB;    //存放矩阵Bint** 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：

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。

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<iostream>
#include<vector>
using namespace std;void GameTable(vector<vector<int> > &vec);
void start();xunhuan.cpp
#include "xunhuan.h"void GameTable(vector<vector<int> > &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<vector<int> > vec(s, vector<int>(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

有兴趣可以加群一起讨论：948097478