实验4：递归与分治策略的应用

课程实验报告结果运行到一定规模：快速排序：运行到一定规模：堆排序：运行到一定规模：矩阵乘法：1朴素算法：2Strassen矩阵乘算法一定规模后：总结横坐标计算规模：1：8129 2：65536 3：131072 4：262144 5：1048576随着输入规模的增大，通过三种算法的时间记录做成折线图观察不难发现，在初期，三种算法所用时间几乎相等，随着输入规模的不断增大，堆排序和快速排序仍然能够保持相对较小的增长，而并归排序所用时间复杂度开始大幅度增加。

快速排序果然是快，数据越大优势越明显，并且实现上也较为简单。

理论上它的平均时间和归并排序，堆排序都是一样的（在最坏情况还还不如它们），都是O(nlog2n)，但实际运行来看比它们两者的速度都快一倍以上。

COOL！合并排序需要额外相同规模的数组，空间复杂度为O（n）。

从具体实现来看，这只是一种理论上的优秀算法，想法比较简单直接，但实现上比quicksort 复杂，运行时间也差，在数据很大的时候运行时间是heapsort的两倍，更不用说quicksort了。

堆排序利用了二分树的结构，将时间复杂度降到O(nlog2n),理论上和实现上表现都不错,并且发现在数据量是10 000 000时，甚至优于快排，可能是运行时数据的问题。

对于strassen 算法对其时间复杂度分析：T（n）=7T(n/2)+O(n);而朴素算法的时间复杂度为n的三次方。

随着数据增大，也出现乘方级别的时间复杂度差距。

附录//头文件#include<iostream>#include<stdio.h>#include<windows.h>#include<time.h>#include<string.h>#define PARENT(i) (i/2) //几个较简单函数#define LEFT(i) (2*i+1)#define RIGHT(i) (2*i+2)using namespace std;//定义所需要变量等#define MAX 100000int a[MAX]; //数组存储原始顺序int temp[MAX]; //临时数组存储临时排序值int num; //计算统计逆序对数int N = 2; //数据规模clock_t begintimes, endtimes; //clock_t为clock()函数返回的变量类型double duration; //运行时间计算int heapsize; //堆长度//随机生成数函数int number(){int a;a = rand() % 10000 + 1; //随机生成1到一万之间的整数return a;}//初始化函数对数组a[]初始化。

void init(){memset(temp,0, MAX * sizeof(int)); //临时数组清零for (int i = 0; i < N; i++){ //新数组赋值a[i] = number();}return;}//单次并归挑选void Merge(int left, int mid, int right) //需要三个参数，将原来数组分割{int i = left, j = mid + 1, n = 0, length = right - left;//i开始为左半部分最左边，j为右半部分最左边while (i <= mid && j <= right){ //未超限进行循环填数if (a[i]>a[j]){ //左边比右边大temp[n++] = a[j++];num += mid - i + 1; //从i到mid都比a[j]大}else{temp[n++] = a[i++];}}if (i>mid){ //左边全部填满了，填右边while (j <= right){temp[n++] = a[j++];}}else{ //右边填满，填左边while (i <= mid){temp[n++] = a[i++];}}for (int k = 0; k <= length; k++){ //最后临时数组赋值到原数组a[left + k] = temp[k];}//递归进行并归排序void MergeSort(int left, int right){if (left<right){int mid = (left + right) / 2;MergeSort(left, mid);MergeSort(mid + 1, right);Merge(left, mid, right);}}//快速排序一次int Partition(int left, int right){int i = left - 1;for (int j = left; j <= right - 1; j++){if (a[j] < a[right]){ //把right作为轴i++; //这个i坐标左边的值是比a[right]小的swap(a[i], a[j]); //交换}}swap(a[i + 1], a[right]); //最后把i+1和right交换，这样轴就是i+1了必须是保证i+1上当初就是作为标杆的a[right]啊。

return i + 1;}//递归进行快排整体void QuickSort(int left, int right){if (left<right){int q = Partition(left, right);QuickSort(left, q - 1);QuickSort(q + 1, right);}}//堆排序，函数太多，新建一个命名空间namespace MySort{template<typename T> //堆排序的大顶堆优化（找数）void Max_Heapify(T*arr, int i, size_t heapSize){//从元素A[i]、A[LEFT(i)]、A[RIGHT(i)]中找出最大的，并将其下标保存在Largest中//size_t heapSize = sizeof(arr) / sizeof(*(arr)); 也就是数量nint l = LEFT(i);int r = RIGHT(i);int largest;//寻找if (l<heapSize && *(arr + l)>*(arr + i))largest = l;elselargest = i;if (r<heapSize && *(arr + r)>*(arr + largest))largest = r;if (largest != i){swap(*(arr + i), *(arr + largest));Max_Heapify(arr, largest, heapSize);}//如果A[i]是最大的，则以i为根的子树已经是最大堆}template<typename T> //建立大顶堆，采用上面大顶堆方法进行优化void Build_Max_Heap(T*arr, size_t heapSize){ //从底部开始进行向上优化for (int i = heapSize / 2 - 1; i >= 0; i--)Max_Heapify(arr, i, heapSize);}template <typename T> //获得最大顶堆，堆排序开始，即元素出堆void HeapSort(T *arr, size_t heapSize){Build_Max_Heap(arr, heapSize);for (int i = heapSize - 1; i > 0; i--){swap(*arr, *(arr + i));Max_Heapify(arr, 0, i);}}}int main(){N = 2;do{N *= 2; //依次增大计算规模srand((unsigned)time(NULL)); //给一个时间种子init();//初始化一次cout << "进行规模为" << N << "的排序" << endl;cout << "原始数组为：";for (int i = 0; i < N; i++){cout << a[i] << " ";}cout << endl;begintimes = clock(); //计时开始MergeSort(0,N-1);QuickSort(0,N-1);MySort::HeapSort<int>(a, N);endtimes = clock(); //计时结束duration = 1000 * (double)(endtimes - begintimes) / CLK_TCK; //总共用时(毫秒)cout << "排序后数组为：";for (int i = 0; i < N; i++){cout << a[i] << " ";}cout << endl;cout << "此次用时为" << duration << "毫秒" << endl << endl << endl;//记录实验结果，注意运行一次手动进行数据转移，清除数据FILE *fpWrite1 = fopen("data1.txt", "a+"); //记录实验结果fprintf(fpWrite1, "%d\n", N);fclose(fpWrite1);FILE *fpWrite2 = fopen("data2.txt", "a+"); //记录实验结果fprintf(fpWrite2, "%d\n", duration);fclose(fpWrite2);} while (duration < 180000); //单次时间小于3分钟return 0;}#include<iostream>#include<stdio.h>#include<time.h>#include<windows.h>#define MAX 10000using namespace std;int N;clock_t begintimes, endtimes; //clock_t为clock()函数返回的变量类型double duration; //运行时间计算//随机生成数函数int number(){int a;a = rand() % 100 + 1; //随机生成1到一万之间的整数return a;}//最朴素算法三重循环void pusu(int **arr, int **brr, int **crr){for (int i = 0; i <= N - 1; i++) {for (int j = 0; j <= N - 1; j++) {for (int k = 0; k <= N - 1; k++) {crr[i][j] += arr[i][k] * brr[k][j];}}}}//Strassen矩阵乘法算法，矩阵分块，仅仅针对2的n次幂次阶处理void gerResultStrassen(int **arr, int **brr, int n, int **crr){ if (n == 1){crr[0][0] += arr[0][0] * brr[0][0];}else{int m = n / 2;int **arr11 = new int*[m];int **arr12 = new int*[m];int **arr21 = new int*[m];int **arr22 = new int*[m];int **brr11 = new int*[m];int **brr12 = new int*[m];int **brr21 = new int*[m];int **brr22 = new int*[m];int **crr11 = new int*[m];int **crr12 = new int*[m];int **crr21 = new int*[m];int **crr22 = new int*[m];for (int i = 0; i < m; ++i){arr11[i] = new int[m];arr12[i] = new int[m];arr21[i] = new int[m];arr22[i] = new int[m];brr11[i] = new int[m];brr12[i] = new int[m];brr21[i] = new int[m];brr22[i] = new int[m];crr11[i] = new int[m];crr12[i] = new int[m];crr21[i] = new int[m];crr22[i] = new int[m];}//获取矩阵//四块矩阵的分别计算//11for (int i = 0; i < m; ++i){for (int j = 0; j < m; ++j){arr11[i][j] = arr[i][j];brr11[i][j] = brr[i][j];}}//22for (int i = m; i < n; ++i){for (int j = m; j < n; ++j){arr22[i - m][j - m] = arr[i][j];brr22[i - m][j - m] = brr[i][j];}}//12for (int i = 0; i < m; ++i){for (int j = m; j < n; ++j){arr12[i][j - m] = arr[i][j];brr12[i][j - m] = brr[i][j];}}//21for (int i = m; i < n; ++i){for (int j = 0; j < m; ++j){arr21[i - m][j] = arr[i][j];brr21[i - m][j] = brr[i][j];}}for (int i = 0; i < m; ++i){for (int j = 0; j < m; ++j){crr11[i][j] = 0;crr12[i][j] = 0;crr21[i][j] = 0;crr22[i][j] = 0;}}//递归分治gerResultStrassen(arr11, brr11, m, crr11); gerResultStrassen(arr12, brr21, m, crr11);gerResultStrassen(arr11, brr12, m, crr12);gerResultStrassen(arr12, brr22, m, crr12);gerResultStrassen(arr21, brr11, m, crr21); gerResultStrassen(arr22, brr21, m, crr21);gerResultStrassen(arr21, brr12, m, crr22); gerResultStrassen(arr22, brr22, m, crr22); //一下是矩阵的分为四块//11for (int i = 0; i < m; ++i){for (int j = 0; j < m; ++j){crr[i][j] += crr11[i][j];}}//22for (int i = m; i < n; ++i){for (int j = m; j < n; ++j){crr[i][j] += crr22[i - m][j - m];}}//12for (int i = 0; i < m; ++i){for (int j = m; j < n; ++j){crr[i][j] += crr12[i][j - m];}}//21for (int i = m; i < n; ++i){for (int j = 0; j < m; ++j){crr[i][j] += crr12[i - m][j];}}//后期处理for (int i = 0; i < m; ++i){delete[] arr11[i];delete[] brr11[i];delete[] crr11[i];delete[] arr12[i];delete[] brr12[i];delete[] crr12[i];delete[] arr21[i];delete[] brr21[i];delete[] crr21[i];delete[] arr22[i];delete[] brr22[i];delete[] crr22[i];}delete[] arr11;delete[] brr11;delete[] crr11;delete[] arr12;delete[] brr12;delete[] crr12;delete[] arr21;delete[] brr21;delete[] crr21;delete[] arr22;delete[] brr22;delete[] crr22;}}//初始化函数void init(int **arr, int **brr, int **crr){//初始化赋值for (int i = 0; i < N; ++i){for (int j = 0; j < N; ++j){arr[i][j]=number();crr[i][j] = 0;}}for (int i = 0; i < N; ++i){for (int j = 0; j < N; ++j){brr[i][j]=number();}}}//输出函数void input(int **arr, int **brr, int **crr){ cout << "矩阵A\n";for (int i = 0; i < N; ++i){for (int j = 0; j < N; ++j){cout << arr[i][j] << " ";}cout << endl;}cout << "矩阵B\n";for (int i = 0; i < N; ++i){for (int j = 0; j < N; ++j){cout << brr[i][j] << " ";}cout << endl;}cout << "相乘后的矩阵C\n";for (int i = 0; i < N; ++i){for (int j = 0; j < N; ++j){cout << crr[i][j] << " ";}cout << endl;}cout << "所用时间为：" << duration << endl;}//主函数int main(){N = 2;//矩阵的阶数do{N = N * 2;int **arr = new int*[N]; //定义数组，分别存放三个矩阵int **brr = new int*[N];int **crr = new int*[N];//获取矩阵for (int i = 0; i < N; ++i){arr[i] = new int[N];brr[i] = new int[N];crr[i] = new int[N];。