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#include <stdio.h>
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#include <ctime>
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#define MAX 1024
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//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
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void sparse_matmul_coo(float* A_values,int* A_rowIndex, int* A_colIndex, int A_nonZeroCount,
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float* B_values, int* B_rowIndex, int* B_colIndex,int B_nonZeroCount,
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float* C_values, int* C_rowIndex, int* C_colIndex,int* C_nonZeroCount)
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{
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int currentIndex = 0;
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//遍历 A 非零元素
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for (int i =0 ; i < A_nonZeroCount; i++)
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{
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int rowA = A_rowIndex[i];
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int colA = A_colIndex[i];
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float valueA = A_values[i];
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// 遍历 B 的非零元素
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for (int j =0 ; j < B_nonZeroCount; j++)
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{
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int rowB = B_rowIndex[j];
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int colB = B_colIndex[j];
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float valueB = B_values[j];
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// 如果 A 的列和 B 的行匹配，则计算乘积并存储结果
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if (colA == rowB)
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{
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float product = valueA * valueB;
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// 检查是否已有此 (rowA，colB) 项
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int found = 0;
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for (int k =0; k < currentIndex; k++)
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{
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if (C_rowIndex[k] == rowA && C_colIndex[k] == colB)
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{
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C_values[k] += product;
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found = 1;
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break;
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}
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}
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// 如果没有此项，添加新的非零元素
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if (!found)
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{
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C_values[currentIndex] = product;
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C_rowIndex[currentIndex] = rowA;
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C_colIndex[currentIndex]= colB;
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currentIndex++;
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}
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}
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}
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}
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//更新非零元素数量
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*C_nonZeroCount=currentIndex;
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}
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int main()
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{
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//矩阵 A 的 COO 格式
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float A_values[] = {1, 2, 3,4,5};
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int A_rowIndex[] = {0, 0, 1, 2, 2};
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int A_colIndex[] = {0, 2, 1,0, 2};
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int A_nonZeroCount = 5;
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// 矩阵 B 的 COO 格式
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float B_values[] = {6,8,7,9};
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int B_rowIndex[] = {0,2, 1, 2};
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int B_colIndex[] ={0,0,1, 2};
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int B_nonZeroCount=4;
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// 结果矩阵 的 Coo 格式
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float C_values[MAX];
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int C_rowIndex[MAX];
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int C_colIndex[MAX];
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int C_nonZeroCount =0 ;
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clock_t start = clock();
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sparse_matmul_coo(A_values,A_rowIndex,A_colIndex,A_nonZeroCount,
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B_values,B_rowIndex,B_colIndex,B_nonZeroCount,
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C_values,C_rowIndex,C_colIndex,&C_nonZeroCount) ;
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clock_t end = clock();
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int rowA=0;int colA=0;
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int rowB=0;int colB=0;
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for(int i=0;i<sizeof(A_rowIndex) / sizeof(A_rowIndex[0]);i++)
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{
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if(A_rowIndex[i]>rowA)
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{
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rowA=A_rowIndex[i];
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}
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}
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for(int i=0;i<sizeof(A_colIndex) / sizeof(A_colIndex[0]);i++)
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{
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if(A_colIndex[i]>colA)
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{
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colA=A_colIndex[i];
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}
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}
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for(int i=0;i<sizeof(B_rowIndex) / sizeof(B_rowIndex[0]);i++)
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{
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if(B_rowIndex[i]>rowB)
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{
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rowB=B_rowIndex[i];
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}
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}
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for(int i=0;i<sizeof(B_colIndex) / sizeof(B_colIndex[0]);i++)
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{
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if(B_colIndex[i]>colB)
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{
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colB=B_colIndex[i];
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}
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}
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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);
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// 计算并输出基础的稀疏矩阵乘法的时间
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double coo_time_spent = double(end - start) / CLOCKS_PER_SEC;
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printf("基础的稀疏矩阵乘法时间：%lf秒\n", coo_time_spent);
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} |
