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。
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