dijkstra算法在基于堆的优先级队列中的查找时间

Question

I want to use priority_queue as vertex container for a dikjstra algorithm, after calling extractMin to get a vertex u , I find all adjacent vertex v , and then I might call decreaseKey for v , I known that the decreaseKey spend O(lgN) time, however, before calling decreaseKey , I have to first find the position of v .

I used std::vector and std::make_heap/std::push_heap/std::pop_heap to maintain the priority_queue, using this data structure, to find a specific data would spend O(N) time, which would make the O(lgN) decreaseKey meaningless.

So, what is the common method for the vertex container in dikjstra algorithm, or should I add a member in class Vertex to keep its position in the heap?

Answer 1

First of all you do not need to use std::***_heap functions; There already is priority_queue in STL.

As for updating values already in the heap, you may insert pair s of index and distance. And maintain a distance vector to verify if the distance is still valid or has got updated; something like:

typedef struct {
    size_t index;   /* index of vertex and the distance at which */
    int dist;       /* the vertex was pushed to the heap         */
} value_t;

/* `a` has less priority than `b` if it is further away */
const auto comp = [](const value_t & a, const value_t & b){
    return b.dist < a.dist;
};

priority_queue<value_t, vector<value_t>, decltype(comp)> heap(comp);

/* the *most current* shortest distances for all nodes */
vector<int> dist(n, inf); 

Then the Dikjstra loop would be something like:

while(!heap.empty()) {
    const auto top = heap.top();
    heap.pop();

    if(dist[top.index] < top.dist) {
          /* the node has already been added at a  */
          /* shorter distance therefore skip it    */
          continue;
    }
    /* process the node & its neighbors */
    /* push to the heap neighbors which get their `dist` deccreased */
}

It is true that there may be multiple copies of a same node in the heap (at different distances where only one of them is still valid); but you may show that the size of the heap is O(num. of edges) , therfore the heap would still perform O(log num. of nodes) .