我正在尝试运行我在此处下载的 Dijkstra 算法的 openmp 实现 heather.cs.ucdavis.edu/~matloff/OpenMP/Dijkstra.c
例如,如果我从 5 到 6 再添加一个顶点,以便从第 0 开始的路径经过两个顶点,我的程序无法给我正确的结果,说第 0 和第 6 之间的距离是无限的:^( 可能是什么原因?
#define LARGEINT 2<<30-1 // "infinity"
#define NV 6
// global variables, all shared by all threads by default
int ohd[NV][NV], // 1-hop distances between vertices
mind[NV], // min distances found so far
notdone[NV], // vertices not checked yet
nth, // number of threads
chunk, // number of vertices handled by each thread
md, // current min over all threads
mv; // vertex which achieves that min
void init(int ac, char **av)
{ int i,j;
for (i = 0; i < NV; i++)
for (j = 0; j < NV; j++) {
if (j == i) ohd[i][i] = 0;
else ohd[i][j] = LARGEINT;
}
ohd[0][1] = ohd[1][0] = 40;
ohd[0][2] = ohd[2][0] = 15;
ohd[1][2] = ohd[2][1] = 20;
ohd[1][3] = ohd[3][1] = 10;
ohd[1][4] = ohd[4][1] = 25;
ohd[2][3] = ohd[3][2] = 100;
ohd[1][5] = ohd[5][1] = 6;
ohd[4][5] = ohd[5][4] = 8;
for (i = 1; i < NV; i++) {
notdone[i] = 1;
mind[i] = ohd[0][i];
}
}
// finds closest to 0 among notdone, among s through e
void findmymin(int s, int e, int *d, int *v)
{ int i;
*d = LARGEINT;
for (i = s; i <= e; i++)
if (notdone[i] && mind[i] < *d) {
*d = ohd[0][i];
*v = i;
}
}
// for each i in [s,e], ask whether a shorter path to i exists, through
// mv
void updateohd(int s, int e)
{ int i;
for (i = s; i <= e; i++)
if (mind[mv] + ohd[mv][i] < mind[i])
mind[i] = mind[mv] + ohd[mv][i];
}
void dowork()
{
#pragma omp parallel // Note 1
{ int startv,endv, // start, end vertices for this thread
step, // whole procedure goes NV steps
mymd, // min value found by this thread
mymv, // vertex which attains that value
me = omp_get_thread_num(); // my thread number
#pragma omp single // Note 2
{ nth = omp_get_num_threads(); chunk = NV/nth;
printf("there are %d threads\n",nth); }
// Note 3
startv = me * chunk;
endv = startv + chunk - 1;
for (step = 0; step < NV; step++) {
// find closest vertex to 0 among notdone; each thread finds
// closest in its group, then we find overall closest
#pragma omp single
{ md = LARGEINT; mv = 0; }
findmymin(startv,endv,&mymd,&mymv);
// update overall min if mine is smaller
#pragma omp critical // Note 4
{ if (mymd < md)
{ md = mymd; mv = mymv; }
}
// mark new vertex as done
#pragma omp single
{ notdone[mv] = 0; }
// now update my section of ohd
updateohd(startv,endv);
#pragma omp barrier
}
}
}
int main(int argc, char **argv)
{ int i;
init(argc,argv);
dowork();
// back to single thread now
printf("minimum distances:\n");
for (i = 1; i < NV; i++)
printf("%d\n",mind[i]);
}
最佳答案
这里有两个问题:
如果线程数不能均分value的个数,那么这种分工
startv = me * chunk;
endv = startv + chunk - 1;
将保留最后一个 (NV - nth*(NV/nth))
元素未完成,这意味着距离保留在 LARGEINT
。这可以通过多种方式解决;目前最简单的方法是将所有剩余工作交给最后一个线程
if (me == (nth-1)) endv = NV-1;
(这会导致不必要的负载不平衡,但这是让代码正常工作的合理开始。)
另一个问题是在设置 notdone[]
#pragma omp barrier
#pragma omp single
{ notdone[mv] = 0; }
这确保 notdone
被更新并且 updateohd()
只有在每个人都完成了他们的 findmymin()
并更新了 之后才开始md
和 mv
。
请注意,很容易在您开始使用的原始代码中引入错误;使用的全局变量使其很难推理。 John Burkardt 有一个 nicer version在他的网站上使用相同的算法进行教学,评论几乎非常好,更容易追踪。
关于c++ - Dijkstra 的算法 openmp 奇怪的行为,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13438704/