Question

   for(int k=1; k<=Vertices-1; k++){   
        /*Every sublist has found the shortest to its adjacent vertices
        thats why starting loop from k-1 then going into every edge of a vertex
        and updating the shortest distance from it to the other.
        */
        for(int i=k-1; i<Vertices; i++){
            // Visiting every Vertex(V) and checking distance of its edges to some other vertex Vo.
            for(int j=0; j<Edges[i].size(); j++){
                int v = Edges[i].get(j).src;
                int edge = Edges[i].get(j).dest;
                int weight = Edges[i].get(j).weight;
                // if exisiting distance is not infinity and from source to that vertex the weight is less then update it
                if (dist[v]!=Integer.MAX_VALUE && dist[v]+weight<dist[edge]){
                    dist[edge]=dist[v]+weight;
                    //updating parent of destination to source.
                    parent[edge] = v;
                }
            }
        }
    }

我已经从（LinkedList）列表中实现了bellman ford，因为算法运行V-1（循环1）次并进入每个顶点（在Loop2中）它检查所有边缘（在loop3中）并更新距离目的地。我在这里感到困惑的是，时间复杂度仍然是O（VE）或改变，我已经看到这个工作在2个循环中完成，这就是为什么，并且也将找到每个顶点经过的最短路径或者我必须从0开始吗？

Answer 1

就V时间检查所有边缘而言，复杂度仍为O(VE)。有关详细信息，请阅读this。

贝尔曼福特表演

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