dijkstra | 易学教程

Why doesn't Dijkstra's algorithm work for negative weight edges?

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问题 Can somebody tell me why Dijkstra's algorithm for single source shortest path assumes that the edges must be non-negative. I am talking about only edges not the negative weight cycles. 回答1: Recall that in Dijkstra's algorithm, once a vertex is marked as "closed" (and out of the open set) - the algorithm found the shortest path to it , and will never have to develop this node again - it assumes the path developed to this path is the shortest. But with negative weights - it might not be true.

Dijkstra(迪杰斯特拉求最短路径)-01-有向图-网络延迟时间

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有 N 个网络节点，标记为 1 到 N。给定一个列表 times，表示信号经过有向边的传递时间。 times[i] = (u, v, w)，其中 u 是源节点，v 是目标节点， w 是一个信号从源节点传递到目标节点的时间。现在，我们向当前的节点 K 发送了一个信号。需要多久才能使所有节点都收到信号？如果不能使所有节点收到信号，返回 -1。注意: N 的范围在 [1, 100] 之间。 K 的范围在 [1, N] 之间。 times 的长度在 [1, 6000] 之间。所有的边 times[i] = (u, v, w) 都有 1 <= u, v <= N 且 0 <= w <= 100。 1 class Solution { 2 public: 3 int networkDelayTime(vector<vector<int>>& times, int N, int K) { 4 vector<vector<int>>G(N,vector<int>(N,1000)); 5 vector<int>dist(N,1000); 6 for(int i = 0;i<times.size();i++){ 7 G[times[i][0]-1][times[i][1]-1] = times[i][2]; 8 } 9 /////////////////////////////////////

最短路径：单源最短路径与差分约束系统

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学习自《算法导论》和Wiki 无权/权=1：BFS ——O(V+E) 有向图(非负边)：Dijkstra——O(E+VlgV) 有向图(无环)：Dijkstra——O(V+E) 一般(可正可负可环)：Bellman-Ford———O(VE) 优化：SPFA 一丶BFS BFS ( G , s ) { //初始化 foreach ( v∈G . V ) { v . visit = 0 ; u . d = NIF ; u . path = nil } s . d = 0 ; s . path = nil ; s . visit = 1 ; //根节点进队 enqueue ( Q , s ) while ( Q not empty ) { u = dequeue ( Q ) foreach ( v∈Adj [ u ] ) if ( visit [ v ] == 0 ) { d . v = d . u + 1 ; v . path = u ; v . visit = 1 ; enqueue ( Q , v ) ; } } 复杂度分析 1.初始化成本O(V) 2.队列操作总时间为O(V) 3.每个结点进队和出队一次，只有出队时候才进行扫描，每个邻接链表最多扫描一次，所以用于扫描邻接表的总时间为O(E) 所以O(V+E) 松弛操作每个结点v来说，我们维持一个属性v.d

（最短路dijkstra)B-旅行

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B-旅行题目描述：小z放假了，准备到RRR城市旅行，其中这个城市有N个旅游景点。小z时间有限，只能在三个旅行景点进行游玩。小明租了辆车，司机很善良，说咱不计路程，只要你一次性缴费足够，我就带你走遍RRR城。小z很开心，直接就把钱一次性缴足了。然而小z心机很重，他想选择的路程尽量长。然而司机也很聪明，他每次从一个点走到另外一个点的时候都走最短路径。你能帮帮小z吗？需要保证这三个旅行景点一个作为起点，一个作为中转点一个作为终点。（一共三个景点，并且需要保证这三个景点不能重复）. 输入描述: 本题包含多组输入，第一行输入一个整数t，表示测试数据的组数每组测试数据第一行输入两个数N,M表示RRR城一共有的旅游景点的数量，以及RRR城中有的路的数量。接下来M行，每行三个数，a，b，c表示从a景点和b景点之间有一条长为c的路 t<=40 3<=N,M<=1000 1<=a,b<=N 1<=c<=100 输出描述: 每组数据输出两行，每组数据包含一行，输出一个数，表示整条路程的路长。如果找不到可行解，输出-1. 输入 4 7 7 1 2 100 2 3 100 1 4 4 4 5 6 5 6 10 1 6 4 6 7 8 7 3 1 2 1 1 3 1 1 3 2 7 3 1 2 1 3 4 1 5 6 1 8 9 1 2 1 2 3 1 3 4 1 4 1 1 4 5 1

Shortest route modification

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问题 Is there a way to modify this to show the route of the shortest path? For example, if i had a list of numbers like (3,1),(3,0),(4,3),(2,1) the output for getting from 4 to 1 would be 4->3,3->1 // Prints shortest paths from src to all other vertices void Graph::shortestPath(int src) { // Create a priority queue to store vertices that // are being preprocessed. This is weird syntax in C++. // Refer below link for details of this syntax // http://geeksquiz.com/implement-min-heap-using-stl/

Optimisation of a Dijkstra Shortest Path Search in Delphi

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问题 I'm looking for advices to speed up my implementation of Dijkstra Shortest Path Search on a weighted graph which is a square matrix N x N. The weight on horizontal vertice is called H (resp. V on vertical ones). A picture is worth a thousand words: (source: free.fr) Of course, this is part of a bigger application, but I've extracted the relevant bit here: unit Unit1; interface uses Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms, Dialogs, StdCtrls; const N = 200; /

Dijktra algorithm vs breath first search for shortest path in graph

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问题 I need some clarifications and inputts regarding Dijktra's algorithm vs breath first search in directed graphs, if these are correct. Dijktra's algorithm finds the shortest path from Node A to Node F in a weighted graph regardless of if there is a cycle or not (as long as there are no negative weights) but for that, All paths from A to all other Nodes in the graph are calculated and we grab the path from A to F by reversing the sequences of nodes in prev`. BFS: finds the shortest path from

【图】Dijstra算法

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Dijkstra 算法是典型最短路算法，用于计算一个节点到其他所有节点的最短路径。主要特点是以起始点为中心向外层层扩展，直到扩展到终点为止。代码实现： //Dijkstra bool S[maxsize];//顶点集 int D[maxsize];//到各个顶点的最短路径 int P[maxsize];//记录前驱 void Dijkstra(Graph G,int v){ //初始化 int n=G.vexnum; for(int i=0;i<n;i++){ S[i]=false; D[i]=G.Edge[v][i]; if(D[i]<InF) P[i]=v; else P[i]=-1; } S[v]=true;D[i]=0; //求出最短路径，并入S集 for(int i=1;i<n;i++){ int k,min=INF; for(int j=0;j<n;j++){ if(!S[j]&&D[j]<min){//当前节点未加入S集合且为当前路径最短 min=D[j]; k=j; } } S[k]=true; //更新从源点出发至其余点的最短路径通过K点 for(int j=0;j<n;j++){ if(!S[j]&&D[k]+G.Edge[k][j]<D[j]){ D[j]=D[k]+G.Edge[k][j]<D[j]; P[j]=k; } } } } 来源： CSDN

Netlogo - find the closest agent based on Manhattan distance

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问题 I am modelling a big warehouse operations (see below pic). I have implemented a vertex on each patch (green dots) and link them so that I could compute the shortest path (using dijkstra algorithm) from each vertex to all other vertices, stored in a table (dictionary) of each vertex. This process is done during setup stage and is quite time-consuming. Then the yellow inventory cells (rectangle) will issue the demand request for task fulfillment. The forklift truck (some in the path) is

Fastest algorithm to detect if there is negative cycle in a graph

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问题 I use a matrix d to present a graph. d.(i).(j) means the distance between i and j ; v denotes the number of nodes in the graph. It is possible that there is negative cycle in this graph. I would like to check if a negative cycle exists. I have written something as follows from a variation of Floyd-Warshall: let dr = Matrix.copy d in (* part 1 *) for i = 0 to v - 1 do dr.(i).(i) <- 0 done; (* part 2 *) try for k = 0 to v - 1 do for i = 0 to v - 1 do for j = 0 to v - 1 do let improvement = dr.

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