這是我實現衆所周知的Dijkstra算法的: std::vector<unsigned> RouteFinder::findPathBetweenIntersections(unsigned intersect_id_start, unsigned intersect_id_end) {
//Get number of intersections and reference graph
我發現類似的電影有以下的Cypher查詢: MATCH (m:Movie)-[r*1..2]-(m2:Movie)
WHERE m.movieID = '1'
UNWIND r AS rels
WITH count(rels) as foo, m2, m
ORDER BY foo desc
RETURN DISTINCT m2.title
LIMIT 25
基本上找到有共同關係的電