我曾尝试在不使用优先级队列(堆)的情况下在循环加权图上使用Djikstra的算法,并且该算法有效。
I have tried using Djikstra's Algorithm on a cyclic weighted graph without using a priority queue (heap) and it worked.
维基百科指出,该算法的原始实现不使用优先级队列,并且运行时间为O(V 2 )。
Wikipedia states that the original implementation of this algorithm does not use a priority queue and runs in O(V2) time.
现在,如果我们只是删除优先级队列并使用普通队列,运行时间是线性的,即O(V + E)。
Now if we just removed the priority queue and used normal queue, the run time is linear, i.e. O(V+E).
有人可以解释为什么我们需要优先级队列吗?
Can someone explain why we need the priority queue?
推荐答案我有完全相同的疑问,并找到了一个测试用例,其中没有priority_queue的算法不起作用。
I had the exact same doubt and found a test case where the algorithm without a priority_queue would not work.
假设我有一个Graph对象 g ,方法 addEdge(a,b,w),可将顶点 a 中的边添加到顶点 b 重量 w 。
Let's say I have a Graph object g, a method addEdge(a,b,w) which adds edge from vertex a to vertex b with weight w.
现在,让我定义下图:
Now, let me define the following graph :-
Graph g g.addEdge(0,1,5) ; g.addEdge(1,3,1) ; g.addEdge(0,2,2) ; g.addEdge(2,1,1) ; g.addEdge(2,3,7) ;现在,假设我们的队列包含以下顺序的节点 {0,1,2, 3} 因此,首先访问节点0,然后访问节点1。
Now, say our queue contains the nodes in the following order {0,1,2,3 } So, node 0 is visited first then node 1 is visited.
在这个时间点,dist b / w 0和3使用路径 0-> 1-> 3 计算为6,并将1标记为已访问。
At this point of time the dist b/w 0 and 3 is computed as 6 using the path 0->1->3, and 1 is marked as visited.
Now节点使用路径 0-> 2-> 1 来访问2并将dist b / w 0和1更新为值3,但是由于节点1被标记为已访问,则无法更改距离b / w 0和3(使用最佳路径)(`0-> 2-> 1-> 3)为4。
Now node 2 is visited and dist b/w 0 and 1 is updated to the value 3 using the path 0->2->1, but since node 1 is marked visited, you cannot change the distance b/w 0 and 3 which (using the optimal path) (`0->2->1->3) is 4.
,则您的算法会在不使用priority_queue的情况下失败。
So, your algorithm fails without using the priority_queue.
它报告dist b / w 0和3为6,而实际上应为4。
It reports dist b/w 0 and 3 to be 6 while in reality it should be 4.
现在,这是我用于实现算法的代码:-
Now, here is the code which I used for implementing the algorithm :-
class Graph { public: vector<int> nodes ; vector<vector<pair<int,int> > > edges ; void addNode() { nodes.push_back(nodes.size()) ; vector<pair<int,int> > temp ; edges.push_back(temp); } void addEdge(int n1, int n2, int w) { edges[n1].push_back(make_pair(n2,w)) ; } pair<vector<int>, vector<int> > shortest(int source) // shortest path djkitra's { vector<int> dist(nodes.size()) ; fill(dist.begin(), dist.end(), INF) ; dist[source] = 0 ; vector<int> pred(nodes.size()) ; fill(pred.begin(), pred.end(), -1) ; for(int i=0; i<(int)edges[source].size(); i++) { dist[edges[source][i].first] = edges[source][i].second ; pred[edges[source][i].first] = source ; } set<pair<int,int> > pq ; for(int i=0; i<(int)nodes.size(); i++) pq.insert(make_pair(dist[i],i)) ; while(!pq.empty()) { pair<int,int> item = *pq.begin() ; pq.erase(pq.begin()) ; int v = item.second ; for(int i=0; i<(int)edges[v].size(); i++) { if(dist[edges[v][i].first] > dist[v] + edges[v][i].second) { pq.erase(std::find(pq.begin(), pq.end(),make_pair(dist[edges[v][i].first],edges[v][i].first))) ; pq.insert(make_pair(dist[v] + edges[v][i].second,edges[v][i].first)) ; dist[edges[v][i].first] = dist[v] + edges[v][i].second ; pred[i] = edges[v][i].first ; } } } return make_pair(dist,pred) ; } pair<vector<int>, vector<int> > shortestwpq(int source) // shortest path djkitra's without priority_queue { vector<int> dist(nodes.size()) ; fill(dist.begin(), dist.end(), INF) ; dist[source] = 0 ; vector<int> pred(nodes.size()) ; fill(pred.begin(), pred.end(), -1) ; for(int i=0; i<(int)edges[source].size(); i++) { dist[edges[source][i].first] = edges[source][i].second ; pred[edges[source][i].first] = source ; } vector<pair<int,int> > pq ; for(int i=0; i<(int)nodes.size(); i++) pq.push_back(make_pair(dist[i],i)) ; while(!pq.empty()) { pair<int,int> item = *pq.begin() ; pq.erase(pq.begin()) ; int v = item.second ; for(int i=0; i<(int)edges[v].size(); i++) { if(dist[edges[v][i].first] > dist[v] + edges[v][i].second) { dist[edges[v][i].first] = dist[v] + edges[v][i].second ; pred[i] = edges[v][i].first ; } } } return make_pair(dist,pred) ; }预期结果如下:-
使用priority_queue 0 3 2 4
With priority_queue 0 3 2 4
现在使用没有优先级的队列 0 3 2 6
Now using without priority queue 0 3 2 6
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