为什么Dijkstra的算法使用堆(优先级队列)? [英] Why does Dijkstra's Algorithm use a heap (priority queue)?
问题描述
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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