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p8sljnpht 19 hours ago
parent 5d750743a6
commit 93d0eae3ae

@ -0,0 +1,70 @@
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
#define SIZE 1024
void sparce_matmul_coo(float*, int*, int*, int,
float*, int*, int*, int, float*, int*, int*, int*);
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 C 的 Coo 格式
float C_values[SIZE];
int C_rowIndex[SIZE];
int C_colIndex[SIZE];
int C_nonZeroCount = 0;
clock_t start = clock();
sparce_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();
printf("基础的稀疏矩阵乘法时间:%lf\n", (double)(end-start) / CLOCKS_PER_SEC);
}
void sparce_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;
int i, j, k;
int rowA, colA, rowB, colB;
float valueA, valueB, product;
// 遍历 A 的非零元素
for(i=0; i<A_nonZeroCount; i++) {
rowA = A_rowIndex[i];
colA = A_colIndex[i];
valueA = A_values[i];
// 遍历 B 的非零元素
for(j=0; j<B_nonZeroCount; j++) {
rowB = B_rowIndex[j];
colB = B_colIndex[j];
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) {
product = valueA * valueB;
// 检查是否已有此(rowA, colB) 项
int found = 0;
for(k=0; k<currentIndex; k++) {
if(C_rowIndex[k] == rowA && C_colIndex[k] == colB) {
C_values[k] += product;
break;
}
}
if (!found) {
C_values[currentIndex] = product;
C_rowIndex[currentIndex] = rowA;
C_colIndex[currentIndex] = colB;
currentIndex++;
}
}
}
}
*C_nonZeroCount = currentIndex;
}
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