并行计算 利用openmpi实现雅可比迭代法

雅克比迭代法：

这个方法是用来解线性方程的，即给定系数矩阵$A$和右边的列向量$B$，求满足$A\ast X=B$的列向量$X$。此处不讲原理，只给出实现方法。
输入：系数矩阵$A$(保证$A$是严格对角占优的)和列向量$B$。
输出：列向量$X$。
环境：$VS2019$
$tips：$虽然结果是正确的，但是感觉加速比特别低…有懂的大佬可以探讨一下。
$code：$

#include
#include
#include
#include
#include
#include
#include
#include
#define NUM_THREADS 16	//线程个数
#define eps 1e-6	//精度
using namespace std;

const int n = 10000;	//矩阵大小

double matrix[n][n];	//系数矩阵
double x[n];	//待求解
double y[n];	//临时存储结果
double b[n];	//matrix * x = b

void init()		//注意 系数矩阵A要生成严格对角占优的 这样才能保证雅克比迭代法的收敛性
{
	srand(time(0));
	double sum = 0;
	for (int i = 0; i < n; i++)
	{
		sum = 0;
		for (int j = 0; j < n; j++)
		{
			if (j == i)
				continue;
			matrix[i][j] = rand() % 10 - 5;
			sum += abs(matrix[i][j]);
		}
		if (rand() & 1)		//严格对角占优
			matrix[i][i] = sum + 1;
		else
			matrix[i][i] = -sum - 1;
	}
	for (int i = 0; i < n; i++)
	{
		b[i] = rand() % 10 - 5;
		x[i] = 0;
	}
}

int main()
{
	omp_set_num_threads(NUM_THREADS);

	clock_t start = clock();

	init();
	int rownum = n / NUM_THREADS;	//每个线程处理的行数
	rownum = max(rownum, 1);
	int myid;	//线程id
	double Max = 0;
	#pragma omp parallel 
	{
		myid = omp_get_thread_num();	//线程id
		int rowmin = myid * rownum;		//该线程需要计算的行的起点
		int rowmax = rowmin + rownum;
		rowmax = min(rowmax, n);
		if (myid == NUM_THREADS - 1)
			rowmax = n;
		double sum = 0;
		for(;;)
		{
			Max = 0;
			for (int i = rowmin; i < rowmax; i++)
			{
				sum = b[i];
				for (int j = 0; j < n; j++)
				{
					if (j == i)
						continue;
					sum -= matrix[i][j] * x[j];
				}
				y[i] = sum / matrix[i][i];
				Max = max(Max, abs(y[i] - x[i]));	//记录最大差距值
			}
			#pragma omp single
			{
				memcpy(x, y, sizeof(x));
			}
			if (Max <= eps)
				break;
			#pragma omp barrier
			{

			}
			/*cout << Max << endl;*/
		}
	}

	clock_t end = clock();
	//cout << "系数矩阵为:\n";
	//for (int i = 0; i < n; i++)
	//{
	//	for (int j = 0; j < n; j++)
	//		printf("%-5.0f", matrix[i][j]);
	//	printf("\n");
	//}
	//cout << "\n向量为:\n";
	//for (int i = 0; i < n; i++)
	//	printf("%.0f\n", b[i]);
	//cout << "\n结果为:\n";
	//for (int i = 0; i < n; i++)
	//	printf("%.4f\n", x[i]);
	//cout << "\n";
	cout << NUM_THREADS << "个线程耗时: " << end - start << "\n";
	return 0;
}