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现在我想做的是，对于从 V1 到 V2 的每条边，我想设置 V2 到 V1 的距离（D）。如果 D 小于 V2 的当前距离，那么我们希望将 V2 的当前距离设置为 D，并将 V2 的前任设置为 V1。

我已将 V1 声明并初始化为最短距离（这只是初始点），并将其标记为完成。

问题：如何声明 V2 并设置它的距离？

std::list<Edge>* Graph::shortestPath(int fromVertex, int toVertex){
    //initialize distance array set to INFINITY
    //initialize predecceor set to -1
    //initialize  bool done array to false

    std::list<Edge> *listOfEdges = new std::list<Edge>();
    std::list<Edge>::iterator it;
    Edge *edge;

    double *distance = new double [numVertices];
    int *predecessor = new int [numVertices];
    bool *done = new bool [numVertices];

    for(int i =0; i < numVertices; i++){
        distance[i] = INFINITY;
        predecessor[i] = -1;
        done[i] = false;
    }

    distance[fromVertex] = 0;
    predecessor[fromVertex] = UNDEFINED_PREDECESSOR;
    done[fromVertex] = true;


    for(int i =0; i < numVertices; i++){
        if(!done[i] && distance[i] != INFINITY){
            int V1 = getVertexWithSmallestDistanceThatsNotDone(distance, done);//choose smallest distance           
            done[V1] = true;//set vertice to to V1.


            double D = distance[toVertex] + distance[predecessor[toVertex]];
            if(D < distance[toVertex]){
                D = distance[toVertex];
                predecessor[toVertex] = fromVertex;
            }
        }
        return listOfEdges;
    }
}
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1 回答 1

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您正在返回一个指向 std::list 的指针。您通常会在函数中为此结果分配内存

std::list<Edge> *result = new std::list<Edge>();

然后，您将返回此指针

return result

在获取此结果的外部函数中，您需要释放动态分配的内存：

std::list<Edge>* edges = graph.shortestPath(1,5);

//work with edges

delete edges;
edges = NULL;//good practice to mark it as "not poiting to anything valid"
于 2011-05-13T01:56:18.150 回答