【算法导论】--分治策略Strassen算法（运用下标运算）【c++】

由于偷懒不想用泛型，所以直接用了整型来写了一份

①首先你得有一个矩阵的class matrix

②matrix为了方便用下标进行运算，

matrix的结构如图：(我知道我的字丑。。。)

 

 matrix.h代码如下：(个人并不喜欢把代码全写在一块，对于阅读者是相当巨大的负担，其实自己受不了（逃）)

 1 #pragma once
 2 #include<vector>
 3 using namespace std;
 4 class matrix
 5 {
 6 public:
 7     vector<vector<int>> nums;
 8     int x0, y0;
 9 
10     int size;
11     matrix();
12     ~matrix();
13     matrix(int size, int x0, int y0);
14     matrix(matrix input, int x0, int y0, int size);
15     matrix(vector<vector<int>>input);
16     void display();
17     static void matrixmultiinit(matrix &a, matrix &b);
18     static matrix matrixmulti(matrix a, matrix b);
19     static matrix matrixadd(matrix a, matrix b);
20     static matrix matrixsub(matrix a, matrix b);
21 };

 

 

matrix.cpp对类的实现

一，构造，析构函数

 1 matrix::matrix()
 2 {
 3 }
 4 
 5 
 6 matrix::~matrix()
 7 {
 8 }
 9 
10 matrix::matrix(int size, int x0, int y0)
11 {
12     vector<vector<int>> temp(size, *new vector<int>(size));
13     for (int i = 0; i < size; i++)
14         for (int j = 0; j < size; j++)
15             temp[i][j] = 0;
16     this->nums = temp;
17     this->x0 = x0;
18     this->y0 = y0;
19     this->size = size;
20 
21 }
22 
23 matrix::matrix(matrix input, int x0, int y0, int size)
24 {
25     this->nums = input.nums;
26     this->x0 = x0;
27     this->y0 = y0;
28     this->size = size;
29 }
30 
31 matrix::matrix(vector<vector<int>> input)
32 {
33     this->nums = input;
34     this->x0 = 0;
35     this->y0 = 0;
36     this->size = input.size();
37 }

 

二，a+b，a-b实现

 1 matrix matrix::matrixadd(matrix a, matrix b)
 2 {
 3     matrix result(a.size, 0, 0);
 4     for (int i = 0; i < result.nums.size(); i++)
 5         for (int j = 0; j < result.nums.size(); j++)
 6             result.nums[i][j] = a.nums[a.x0 + i][a.y0 + j] + b.nums[b.x0 + i][b.y0 + j];
 7     return result;
 8 }
 9 
10 matrix matrix::matrixsub(matrix a, matrix b)
11 {
12 
13     matrix result(a.size, 0, 0);
14     for (int i = 0; i < result.nums.size(); i++)
15         for (int j = 0; j < result.nums.size(); j++)
16             result.nums[i][j] = a.nums[a.x0 + i][a.y0 + j] - b.nums[b.x0 + i][b.y0 + j];
17     return result;
18 }

 

三，a*b的实现

 1 matrix matrix::matrixmulti(matrix a, matrix b)
 2 {
 3     int n = a.size;
 4     int halfsize =n / 2;
 5     matrix result(n, 0, 0);
 6     if (n == 1)
 7         result.nums[0][0] = a.nums[a.x0][a.y0] * b.nums[b.x0][b.y0];
 8     else
 9     {
10         matrix temps[10];
11         for (int i = 0; i < 10; i++)
12         {
13             temps[i] = *new matrix(halfsize, 0, 0);
14         }
15 
16         //00--  a.x0,a.y0,halfsize
17         //01-- a.x0,a.y0+halfsize,halfsize
18         //10--  a.x0+halfsize,a.y0,halfsize
19         //11--    a.x0+halfsize,a.y0+halfsize,halfsize
20                 
21         //00-- b.x0,b.y0,halfsize
22         //01-- b.x0,b.y0+halfsize,halfsize
23         //10-- b.x0+halfsize,b.y0,halfsize
24         //11--    b.x0+halfsize,b.y0+halfsize,halfsize
25         temps[0] = temps[0].matrixsub(*new matrix(b, b.x0, b.y0 + halfsize, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//01-11 b
26         temps[1] = temps[1].matrixadd(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0, a.y0 + halfsize, halfsize));//00+01 a
27         temps[2] = temps[2].matrixadd(*new matrix(a, a.x0 + halfsize, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//10+11 a
28         temps[3] = temps[3].matrixsub(*new matrix(b, b.x0 + halfsize, b.y0, halfsize), *new matrix(b, b.x0, b.y0, halfsize));//10-00 b
29         temps[4] = temps[4].matrixadd(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//00+11 a
30         temps[5] = temps[5].matrixadd(*new matrix(b, b.x0, b.y0, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//00+11 b
31         temps[6] = temps[6].matrixsub(*new matrix(a, a.x0, a.y0 + halfsize, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//01-11 a
32         temps[7] = temps[7].matrixadd(*new matrix(b, b.x0 + halfsize, b.y0, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//10+11 b
33         temps[8] = temps[8].matrixsub(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0, halfsize));//00-10 a
34         temps[9] = temps[9].matrixadd(*new matrix(b, b.x0, b.y0, halfsize), *new matrix(b, b.x0, b.y0 + halfsize, halfsize));//00+01 b
35 
36 
37         matrix tempp[7];
38         for (int i = 0; i < 7; i++)
39         {
40             tempp[i] = *new matrix(n / 2, 0, 0);
41         }
42         tempp[0] = tempp[0].matrixmulti(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(temps[0], 0, 0, halfsize));
43         tempp[1] = tempp[1].matrixmulti(*new matrix(temps[1], 0, 0,halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));
44         tempp[2] = tempp[2].matrixmulti(*new matrix(temps[2], 0, 0, halfsize), *new matrix(b, b.x0, b.y0, halfsize));
45         tempp[3] = tempp[3].matrixmulti(*new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize), *new matrix(temps[3], 0, 0, halfsize));
46         tempp[4] = tempp[4].matrixmulti(*new matrix(temps[4], 0, 0, halfsize), *new matrix(temps[5], 0, 0, halfsize));
47         tempp[5] = tempp[5].matrixmulti(*new matrix(temps[6], 0, 0, halfsize), *new matrix(temps[7], 0, 0, halfsize));
48         tempp[6] = tempp[6].matrixmulti(*new matrix(temps[8], 0, 0, halfsize), *new matrix(temps[9], 0, 0, halfsize));
49 
50 
51 
52         matrix result00 = result00.matrixadd(tempp[4], tempp[3]);
53         result00 = result00.matrixsub(result00, tempp[1]);
54         result00 = result00.matrixadd(result00, tempp[5]);
55         matrix result01 = result01.matrixadd(tempp[0], tempp[1]);
56         matrix result10 = result10.matrixadd(tempp[2], tempp[3]);
57         matrix result11 = result11.matrixadd(tempp[4], tempp[0]);
58         result11 = result11.matrixsub(result11, tempp[2]);
59         result11 = result11.matrixsub(result11, tempp[6]);
60 
61         if (n == 3) {
62         for(int i=0;i<n/2+1;i++)
63             for (int j = 0; j < n / 2 + 1; j++) {
64             
65             result.nums[i][j]= result00.nums[i][j];
66             result.nums[i][j+n/2+1] = result01.nums[i][j];
67             result.nums[i+n/2+1][j] = result10.nums[i][j];
68             result.nums[i+n/2+1][j+n/2+1] = result11.nums[i][j];
69             }
70         }
71 
72         for(int i=0;i<n/2;i++)
73             for (int j = 0; j < n / 2; j++) {
74             
75             result.nums[i][j]= result00.nums[i][j];
76             result.nums[i][j+n/2] = result01.nums[i][j];
77             result.nums[i+n/2][j] = result10.nums[i][j];
78             result.nums[i+n/2][j+n/2] = result11.nums[i][j];
79             }
80 
81     }
82     return result;
83 }

 

四，防止size%2！=0的处理函数（即矩阵的行列数为奇数）

 1 void matrix::matrixmultiinit(matrix &a, matrix &b) {
 2 
 3     if (a.nums.size() % 2 != 0)
 4     {
 5         for (int i = 0; i < a.nums.size(); i++)
 6             a.nums[i].push_back(0);
 7         for (int i = 0; i < b.nums.size(); i++)
 8             b.nums[i].push_back(0);
 9         a.nums.push_back(*new vector<int>(a.nums[0].size(), 0));
10         b.nums.push_back(*new vector<int>(b.nums[0].size(), 0));
11         a.size++;
12         b.size++;
13     }
14 }

 

五，输出函数（这个读者随意）

 1 void matrix::display()
 2 {
 3     for (int i = 0; i < this->nums.size(); i++) {
 4 
 5         cout << "||";
 6         for (int j = 0; j < this->nums[i].size(); j++) {
 7             cout << this->nums[i][j] << " ";
 8         }
 9         cout << "||" << endl;
10 
11     }
12 }

 

六，测试函数

 1 #include <iostream>
 2 #include"matrix.h"
 3 int main()
 4 {
 5     vector<vector<int>> input = { 
 6         {1,2,3},
 7         {1,2,3},
 8         {1,1,1},
 9     };
10     matrix test0 = * new matrix(input);
11     matrix test1 = * new matrix(input);
12     matrix test2;
13     test2.matrixmultiinit(test0, test1);
14     test2= test2.matrixmulti(test0, test1);
15     test2.display();
16 
17 }

 

本人比较愚笨，耗时一天半才完成，不知道是不是天气热的原因，人太燥了，沉不下心来思考bug。

 

a*b有一点需要注意的是分块的逻辑应该怎么表示，一开始我用了两个顶点来表示一个矩阵的分块，如图：

然后halfsize还得一个一个的算，然后自己敲错的几率还会加大，并且还不一定表示正确每个分块，然后就逼疯自己了。

感觉这两天被这一堆bug都弄自闭了。。。。

幸好还是撑过来了，这算是我啃算法导论的第一个坎吧。

幸好第二天在csdn里面看到了别人怎么分块的。看到了一个变量halfsize，于是才开始改自己matrix的构造。

虽然一开始不愿意，但是改完之后，竟然一次过了！woc！

给我了一个教训，以后能“少”一个变量尽量“少”一个变量。