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75 lines
2.5 KiB
75 lines
2.5 KiB
#include <stdio.h>
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#include <ctime>
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#define SIZE 1024
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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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int currentIndex = 0;
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// 遍历 A 的非零元素
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for (int i = 0; i < A_nonZeroCount; i++) {
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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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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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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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if (C_rowIndex[k] == rowA && C_colIndex[k] == colB) {
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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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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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int i;
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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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//矩阵 C 的 COO 格式
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float C_values[SIZE];
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int C_rowIndex[SIZE];
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int C_colIndex[SIZE];
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int C_nonZeroCount=0;
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//计时并输出
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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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printf("初始稀疏矩阵乘法时间:%f秒\n",(double)(end-start)/CLOCKS_PER_SEC);
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return 0;
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}
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