#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;
}