我一直在尝试使用OpenMP并行化快速排序,但似乎我在这方面做错了,因为使用的线程越多越慢!
我知道总是包含开销,但是巨大列表中线程数量的增加应该使它更快而不慢(我的情况)。
这是代码享受!
#include <omp.h>
double start_time, end_time;
#include <stdio.h>
#define MAXSIZE 10000 /* maximum array size */
#define MAXWORKERS 8 /* maximum number of workers */
int numWorkers;
int size;
int doge[MAXSIZE];
void breakdown(int, int);
/* read command line, initialize, and create threads */
int main(int argc, char *argv[]) {
srand(time(NULL));
int i;
/* read command line args if any */
size = (argc > 1)? atoi(argv[1]) : MAXSIZE;
numWorkers = (argc > 2)? atoi(argv[2]) : MAXWORKERS;
if (size > MAXSIZE) size = MAXSIZE;
if (numWorkers > MAXWORKERS) numWorkers = MAXWORKERS;
for(i = 0;i<size;i++){
doge[i] = 1+rand()%99;
}
omp_set_num_threads(numWorkers);
start_time = omp_get_wtime();
#pragma omp parallel
{
#pragma omp single nowait
{
breakdown(0, size);
}
}
end_time = omp_get_wtime();
for(i = 0;i<size;i++){
printf("%d ", doge[i]);
}
printf("it took %g seconds\n", end_time - start_time);
}
void breakdown(int from, int to){
if(to-from < 2){
return;
}
int left, right, temp;
int i_pivot = from + rand()%(to-from);
int pivot = doge[i_pivot];
left = from;
right = to;
while (left <= right){
if (doge[left] > pivot){
/* swap left element with right element */
temp = doge[left];
doge[left] = doge[right];
doge[right] = temp;
if (right == i_pivot)
i_pivot = left;
right--;
}
else
left++;
}
/* place the pivot in its place (i.e. swap with right element) */
temp = doge[right];
doge[right] = pivot;
doge[i_pivot] = temp;
#pragma omp task
{
breakdown(from, right - 1);
}
#pragma omp task
{
breakdown(right + 1, to);
}
//implicit DOGE
}
我相信我做错了parallalization错误.. 这些行:
#pragma omp parallel
{
#pragma omp single nowait
{
breakdown(0, size);
}
}
和
#pragma omp task
{
breakdown(from, right - 1);
}
#pragma omp task
{
breakdown(right + 1, to);
}
任何帮助都是doge
答案 0 :(得分:1)
你尝试过更大的阵列吗? 10000不算什么,应该立即排序,你需要数百万的数字让它运行至少几秒钟。