Jinhong_Homework3/task5.cpp

#include <stdio.h>
#include <ctime>
 
#define MAX 1024
//因为所有变量都是局部变量，未采用动态分配内存的办法，故本代码文件不手动清理内存 
void sparse_matmul_coo(float* A_values,int* A_rowIndex, int* A_colIndex, int A_nonZeroCount,
					   float* B_values, int* B_rowIndex, int* B_colIndex,int B_nonZeroCount,
					   float* C_values, int* C_rowIndex, int* C_colIndex,int* C_nonZeroCount) 
{
	int currentIndex = 0;
	//遍历 A 非零元素
	for (int i =0 ; i < A_nonZeroCount; i++) 
	{
		int rowA = A_rowIndex[i];
		int colA = A_colIndex[i];
		float valueA = A_values[i];
		// 遍历 B 的非零元素
		for (int j =0 ; j < B_nonZeroCount; j++) 
		{
			int rowB = B_rowIndex[j];
			int colB = B_colIndex[j];
			float valueB = B_values[j];
			// 如果 A 的列和 B 的行匹配，则计算乘积并存储结果
			if (colA == rowB)
			{
				float product = valueA * valueB;
				// 检查是否已有此 (rowA，colB) 项
				int found = 0;
				for (int k =0; k < currentIndex; k++) 
				{
					if (C_rowIndex[k] == rowA && C_colIndex[k] == colB) 
					{
						C_values[k] += product;
						found = 1;
						break;
					}
				}
				// 如果没有此项，添加新的非零元素
				if (!found) 
				{
					C_values[currentIndex] = product;
					C_rowIndex[currentIndex] = rowA;
					C_colIndex[currentIndex]= colB;
					currentIndex++;
				}
			}
		}
	}
	//更新非零元素数量
	*C_nonZeroCount=currentIndex;
}
				
int main()
{	
	//矩阵 A 的 COO 格式
    float A_values[] = {1, 2, 3,4,5};
    int A_rowIndex[] = {0, 0, 1, 2, 2};
    int A_colIndex[] = {0, 2, 1,0, 2};
    int A_nonZeroCount = 5;
    // 矩阵 B 的 COO 格式
    float B_values[] = {6,8,7,9};
    int B_rowIndex[] = {0,2, 1, 2};
    int B_colIndex[] ={0,0,1, 2};
    int B_nonZeroCount=4; 

	// 结果矩阵 的 Coo 格式
	float C_values[MAX];
	int C_rowIndex[MAX];
	int C_colIndex[MAX];
	int C_nonZeroCount =0 ;
	
	clock_t start = clock();
    sparse_matmul_coo(A_values,A_rowIndex,A_colIndex,A_nonZeroCount,
					  B_values,B_rowIndex,B_colIndex,B_nonZeroCount,
					  C_values,C_rowIndex,C_colIndex,&C_nonZeroCount) ;
    clock_t end = clock();

    int rowA=0;int colA=0;
    int rowB=0;int colB=0;
    for(int i=0;i<sizeof(A_rowIndex) / sizeof(A_rowIndex[0]);i++)
    {
        if(A_rowIndex[i]>rowA)
        {
            rowA=A_rowIndex[i];
        }
    }
    for(int i=0;i<sizeof(A_colIndex) / sizeof(A_colIndex[0]);i++)
    {
        if(A_colIndex[i]>colA)
        {
            colA=A_colIndex[i];
        }
    }
    for(int i=0;i<sizeof(B_rowIndex) / sizeof(B_rowIndex[0]);i++)
    {
        if(B_rowIndex[i]>rowB)
        {
            rowB=B_rowIndex[i];
        }
    }    
    for(int i=0;i<sizeof(B_colIndex) / sizeof(B_colIndex[0]);i++)
    {
        if(B_colIndex[i]>colB)
        {
            colB=B_colIndex[i];
        }
    }
    printf("矩阵A的规模为%d*%d,矩阵B的规模为%d*%d\n",rowA+1,colA+1,rowB+1,colB+1);

    // 计算并输出基础的稀疏矩阵乘法的时间
    double coo_time_spent = double(end - start) / CLOCKS_PER_SEC;
    printf("基础的稀疏矩阵乘法时间：%lf秒\n", coo_time_spent);
}