目的是找到路径：a->b->d->e 为最短路径
然后我现在用的
match (a:test{name:“a”}),(b:test{name:“e”}),
p=((a)-[*]-(b))
with p,reduce(s=0,r in relationships§|s+r.distance)AS dist
return p,dist ORDER by dist limit 1
想问一下有没有更优（即时间更短）的查询方法，刚入门，想不太出。
找到了apoc里面的一个函数，如下，感觉好一点吧应该
match (a:test{name:“a”}),(b:test{name:“e”})
CALL apoc.algo.dijkstra(a,b,“rel”,“distance”)yield path as path,weight as weight
return path,weight
