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In the Single-Source Shortest Paths (SSSP) problem, we aim to find the shortest paths weights (and the actual paths) from a particular single-source vertex to all other vertices in a directed weighted graph (if such paths exist).


The SSSP problem is a(nother) very well-known Computer Science (CS) problem that every CS students worldwide need to be aware of and hopefully master.


The SSSP problem has several different efficient (polynomial) algorithms (e.g. Bellman Ford, BFS, DFS, Dijkstra — 2 versions, and/or Dynamic Programming) that can be used depending on the nature of the input directed weighted graph, i.e. weighted/unweighted, with/without (negative weight) cycle, or structurally special (a tree/a DAG).


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SSSP is one of the most frequent graph problem encountered in real-life. Every time we want to move from one place (usually our current location) to another (our destination), we will try to pick a short — if not the shortest — path.


SSSP algorithm(s) is embedded inside various map software like Google Maps and in various Global Positioning System (GPS) tool.


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输入 1: 一个有向加权图 G(V, E), 不一定是连接着的, 其中 V/顶点可用于描述交叉点,交汇点 ,房屋,地标等,还有 E/边可用于描述街道,道路 ,大道,且具有适当的方向和权重/成本。


输入 2: 顾名思义, SSSP 问题有另一个输入:一个源顶点 sV.


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SSSP 问题的目的是找到从 s 到每个顶点 uV 的最短路径权,表示为 δ(s, u) ( δ 发音为 'delta' ) 以及从 su 的实际最短路径。

路径 p 仅仅是沿改路径的边的权重的总和。

ss 的最短路径的权重是微不足道的:0。从 s 到任何不可到达顶点的最短路径也是微不足道的:+∞。

PS: 从 sv 的最短路径的权重,其中 (s, v) ∈ E ,不一定是 w(s, v) 的权重。请参阅下面的几张幻灯片来理解这一点。

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The outputs of all six (6) SSSP algorithms for the SSSP problem discussed in this visualization are these two arrays/Vectors:

  1. An array/Vector D of size V (D stands for 'distance')
    Initially, D[u] = 0 if u = s; otherwise D[u] = +∞ (a large number, e.g. 109)
    D[u] decreases as we find better (shorter) paths
    D[u]δ(s, u) throughout the execution of SSSP algorithm
    D[u] = δ(s, u) at the end of SSSP algorithm
  2. An array/Vector p of size V (p stands for 'parent'/'predecessor'/'previous')
    p[u] = the predecessor on best path from source s to u
    p[u] = NULL (not defined, we can use a value like -1 for this)
    This array/Vector p describes the resulting SSSP spanning tree
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Initially, D[u] = +∞ (practically, a large value like 109) ∀uV\\{s}, but D[s] = D[0] = 0.
Initially, p[u] = -1 (to say 'no predecessor') ∀uV.


Now click Dijkstra(0) — don't worry about the details as they will be explained later — and wait until it is over (approximately 10s on this small graph).


At the end of that SSSP algorithm, D[s] = D[0] = 0 (unchanged) and D[u] = δ(s, u)uV
e.g. D[2] = 6, D[4] = 7 (these values are stored as red text under each vertex).
At the end of that SSSP algorithm, p[s] = p[0] = -1 (the source has no predecessor), but p[v] = the origin of the orange edges for the rest, e.g. p[2] = 0, p[4] = 2.


Thus, if we are at s = 0 and want to go to vertex 4, we will use shortest path 0 → 2 → 4 with path weight 7.

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Some graphs contain negative weight edge(s) (not necessarily cyclic) and/or negative weight cycle(s). For example (fictional): Suppose you can travel forward in time (normal, edges with positive weight) or back in time by passing through time tunnel (special wormhole edges with negative weight), as the example shown above.


On that graph, the shortest paths from the source vertex s = 0 to vertices {1, 2, 3} are all ill-defined. For example 1 → 2 → 1 is a negative weight cycle as it has negative total path (cycle) weight of 15-42 = -27. Thus we can cycle around that negative weight cycle 0 → 1 → 2 → 1 → 2 → ... forever to get overall ill-defined shortest path weight of -∞.


However, notice that the shortest path from the source vertex s = 0 to vertex 4 is ok with δ(0, 4) = -99. So the presence of negative weight edge(s) is not the main issue. The main issue is the presence of negative weight cycle(s) reachable from source vertex s.

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The main operation for all SSSP algorithms discussed in this visualization is the relax(u, v, w(u, v)) operation with the following pseudo-code:

relax(u, v, w_u_v)
if D[v] > D[u]+w_u_v // if the path can be shortened
D[v] = D[u]+w_u_v // we 'relax' this edge
p[v] = u // remember/update the predecessor
// update some other data structure(s) as necessary

For example, see relax(1,2,4) operation on the figure below: relax operation example

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对于指定一个输入图,有两种不同的方法:

  1. 绘制图: 您可以绘制任何有向加权图作为输入图。
  2. 示例图: 您可以从我们选择的示例图列表中进行挑选,以帮助您入门。这些示例图具有不同的特征。
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In this visualization, we will discuss 6 (SIX) SSSP algorithms.


We will start with the O(V×E) Bellman Ford's algorithm first as it is the most versatile (but also the slowest) SSSP algorithm. We will then discuss 5 (FIVE) other algorithms (including two variants of Dijkstra's algorithm) that solve special-cases of SSSP problem in a much faster manner.

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The general purpose Bellman Ford's algorithm can solve all kinds of valid SSSP problem variants (expect one — the one that is ill-defined anyway, to be discussed soon), albeit with a rather slow O(V×E) running time. It also has an extremely simple pseudo-code:

for i = 1 to |V|-1 // O(V) here, so O(V×E×1) = O(V×E)
for each edge(u, v) ∈ E // O(E) here, e.g. by using an Edge List
relax(u, v, w(u, v)) // O(1) here

Without further ado, let's see a preview of how it works on the example graph above by clicking BellmanFord(0) (≈30s, and for now, please ignore the additional loop at the bottom of the pseudo-code).

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Bellman Ford's algorithm can be made to run slightly faster on normal input graph, from the worst case of O(V×E) to just O(k×E) where k is the number of iterations of Bellman Ford's outer loop.


Discussion: How to do this? Is the speed-up significant?

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To convince the worldwide audience that Bellman Ford's algorithm works, let's temporarily move from visualization mode to proof mode for a few slides.


Theorem 1: If G = (V, E) contains no negative weight cycle, then the shortest path p from source vertex s to a vertex v must be a simple path.


Recall: A simple path is a path p = {v0, v1, v2, ..., vk}, (vi, vi+1) ∈ E, ∀ 0 ≤ i ≤ (k-1) and there is no repeated vertex along this path.

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  1. Suppose the shortest path p is not a simple path
  2. Then p must contains one (or more) cycle(s) (by definition of non-simple path)
  3. Suppose there is a cycle c in p with positive weight (e.g. greenbluegreen on the left image) cycle
  4. If we remove c from p, then we will have a shorter 'shortest path' than our shortest path p
  5. A glaring contradiction, so p must be a simple path
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  1. Even if c is actually a cycle with zero (0) total weight — it is possible according to our Theorem 1 assumption: no negative weight cycle (see the same greenbluegreen but on the right image), we can still remove c from p without increasing the shortest path weight of p cycle
  2. In conclusion, p is a simple path (from point 5) or can always be made into a simple path (from point 6)

In another word, shortest path p has at most |V|-1 edges from the source vertex s to the 'furthest possible' vertex v in G (in terms of number of edges in the shortest path — see the Bellman Ford's Killer example above).

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定理 2: 如果 G = (V, E) 不包含负权重循环 , 那么在贝尔曼福特算法终止后,我们将得到 D[v] = δ(s, u), ∀ uV.


为此, 我们将使用归纳证明法 ,以下是验证的起始点:


考虑从源点 s 到顶点 vi 的最短路径 p ,其中 vi 被定义为顶点,从源点 s 到达它的实际最短路径需要 i 跳 (边) 。回想一下定理 1 ,p 将是简单的路径,因为我们有相同的没有负权重循环的假设。

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  1. Initially, D[v0] = δ(s, v0) = 0, as v0 is just the source vertex s
  2. After 1 pass through E, we have D[v1] = δ(s, v1)
  3. After 2 pass through E, we have D[v2] = δ(s, v2)
  4. ...
  5. After k pass through E, we have D[vk] = δ(s, vk)
  6. When there is no negative weight cycle, the shortest path p is a simple path (see Theorem 1), thus the last iteration should be iteration |V|-1
  7. After |V|-1 pass through E, we have D[v|V|-1] = δ(s, v|V|-1), regardless the ordering of edges in E — see the Bellman Ford's Killer example above
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Try running BellmanFord(0) on the 'Bellman Ford's Killer' example above. There are V = 7 vertices and E = 6 edges but the edge list E is configured to be at its worst possible order. Notice that after (V-1)×E = (7-1)*6 = 36 operations (~40s, be patient), Bellman Ford's will terminate with the correct answer and there is no way we can terminate Bellman Ford's algorithm earlier.

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The only input graph that Bellman Ford's algorithm has issue is the input graph with negative weight cycle reachable from the source vertex s.


However, Bellman Ford's can be used to detect if the input graph contains at least one negative weight cycle reachable from the source vertex s by using the corollary of Theorem 2: If at least one value D[v] fails to converge after |V|-1 passes, then there exists a negative-weight cycle reachable from the source vertex s.


Now run BellmanFord(0) on the example graph that contains negative edges and a negative weight cycle. Please concentrate on the loop at the bottom of the pseudo-code.

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Sometimes, the actual problem that we face is not the general form of the original problem. Therefore in this e-Lecture, we want to highlight five (5) special cases involving the SSSP problem. When we encounter any one of them, we can solve it with different and (much) faster algorithm than the generic O(V×E) Bellman Ford's algorithm. They are:

  1. On Unweighted Graphs: O(V+E) BFS,
  2. On Graphs without negative weight: O((V+E) log V) Dijkstra's algorithm,
  3. On Graphs without negative weight cycle: O((V+E) log V) Modified Dijkstra's,
  4. On Tree: O(V+E) DFS/BFS,
  5. On Directed Acyclic Graphs (DAG): O(V+E) Dynamic Programming (DP)
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O(V+E) 广度优先搜索 (BFS) 算法可以解决有特殊情况的SSSP问题,当输入图是未加权时(所有边都具有单位权重1,在上面的例子 'CP3 4.3' 上尝试 BFS(5)),或者正常数加权(所有有边都具有相同的恒定权重,例如,你可以在上面的例图中用你所选择的任何正恒定权重去更改所有的边的权重)。

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When the graph is unweighted — this appears quite frequently in real life — the SSSP problem can be viewed as a problem of finding the least number of edges traversed from the source vertex s to other vertices.


The BFS spanning tree from source vertex s produced by the fast O(V+E) BFS algorithm — notice the + sign — precisely fits the requirement.


Compared with the O(V×E) of Bellman Ford's — notice the × sign — it is a no-brainer to use BFS for this special case of SSSP problem.

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Compared to the standard BFS in Graph Traversal module, we need to perform simple modifications to make BFS able to solve the unweighted version of the SSSP problem:

  1. First, we change the Boolean array visited into an Integer array D.
  2. At the start of BFS, instead of setting visited[u] = false, we set D[u] = 1e9 (a large number to symbolise +∞ or even -1 to symbolise 'unvisited' state, but we cannot use 0 as D[0] = 0) ∀uV\{s}; Then we set D[s] = 0
  3. We change the BFS main loop from
    if (visited[v] = 0) { visited[v] = 1 ... } // v is unvisited
    to
    if (D[v] = 1e9) { D[v] = D[u]+1 ... } // v is 1 step away from u
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However, BFS will very likely produce wrong answer when run on weighted graphs as BFS is not actually designed for to solve the weighted version of SSSP problem. There may be a case that taking a path with more number of edges used produces lower total overall path weight than taking a path with minimum number of edges used — which is the output of BFS algorithm.


In this visualization, we will allow you to run BFS even on 'wrong' input graph for pedagogical purpose, but we will display a warning message at the end of the algorithm. For example, try BFS(0) on the general graph above and you will see that vertices {3,4} will have wrong D[3] and D[4] values (and also p[3] and p[4] values).


We will soon see Dijkstra's algorithm (2 implementation variants) for solving certain weighted SSSP problems in a faster way than the general Bellman Ford's algorithm.

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The O((V+E) log V) Dijkstra's algorithm is the most frequently used SSSP algorithm for typical input: Directed weighted graph that has no negative weight edge at all, formally: ∀ edge(u, v) ∈ E, w(u, v) ≥ 0. Such weighted graph is very common in real life as travelling from one place to another always use positive time unit(s). Try Dijkstra(0) on one of the Example Graphs: CP3 4.17 shown above.

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Dijkstra's algorithm maintains a set S (Solved) of vertices whose final shortest path weights have been determined. Initially S = {s}, the source vertex s only.


Then, it repeatedly selects vertex u in {V\S} with the minimum shortest path estimate, adds u to S, and relaxes all outgoing edges of u.


This entails the use of a Priority Queue as the shortest path estimates keep changing as more edges are processed. The choice of relaxing edges emanating from vertex with the minimum shortest path estimate first is greedy, i.e. use the "best so far", but we will see later that it can be proven that it will eventually ends up with an optimal result — if the graph has no negative weight edge.

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In Dijkstra's algorithm, each vertex will only be extracted from the Priority Queue (PQ) once. As there are V vertices, we will do this maximum O(V) times.


ExtractMin() operation runs in O(log V) whether the PQ is implemented using a Binary Min Heap or using a balanced BST like AVL Tree.


Therefore this part is O(V log V).

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Every time a vertex is processed, we relax its neighbors. In total, E edges are processed.


If by relaxing edge(u, v), we have to decrease D[v], we call the O(log V) DecreaseKey() operation in Binary Min Heap (harder to implement as C++ STL priority_queue/Java PriorityQueue does not support this operation efficiently yet) or simply delete the old entry and then re-insert a new entry in balanced BST like AVL Tree (which also runs in O(log V), but this is much easier to implement, just use C++ STL set/Java TreeSet).


Therefore, this part is O(E log V).


Thus in overall, Dijkstra's algorithm runs in O(V log V + E log V) = O((V+E) log V) time, which is much faster than the O(V×E) Bellman Ford's algorithm.

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When the input graph contains at least one negative weight edge — not necessarily negative weight cycle — Dijkstra's algorithm can produce wrong answer.


Try Dijkstra(0) on one of the Example Graphs: CP3 4.18.


At the end of the execution of Dijkstra's algorithm, vertex 4 has wrong D[4] value as the algorithm started 'wrongly' thinking that subpath 0 → 1 → 3 is the better subpath of weight 1+2 = 3, thus making D[4] = 6 after calling relax(3,4,3). However, the presence of negative weight -10 at edge 2 → 3 makes the other subpath 0 → 2 → 3 eventually the better subpath of weight 10-10 = 0 although it started worse with path weight 10 after the first edge 0 → 2. This better D[3] = 0 is never propagated further due to the greedy nature of Dijkstra's algorithm, hence D[4] is wrong.

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Dijkstra's algorithm can also be implemented differently. The O((V+E) log V) Modified Dijkstra's algorithm can be used for directed weighted graphs that may have negative weight edges but no negative weight cycle.


Such input graph appears in some practical cases, e.g. travelling using an electric car that has battery and our objective is to find a path from source vertex s to another vertex that minimizes overall battery usage. As usual, during acceleration (or driving on flat/uphill road), the electric car uses (positive) energy from the battery. However, during braking (or driving on downhill road), the electric car recharges (or use negative) energy to the battery. There is no negative weight cycle due to kinetic energy loss.


For example, try ModifiedDijkstra(0) on one of the Example Graphs: CP3 4.18 that has troubled the original version of Dijkstra's algorithm (see previous slide).

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The key idea is the modification done to C++ STL priority_queue/Java PriorityQueue to allow it to perform the required 'DecreaseKey' operation efficiently, i.e. in O(log V) time.


The technique is called 'Lazy Update' where we leave the 'outdated/weaker/bigger-valued information' in the Min Priority Queue instead of deleting it straight-away. As the items are ordered from smaller values to bigger values in a Min PQ, we are guaranteeing ourself that we will encounter the smallest/most-up-to-date item first before encountering the weaker/outdated item(s) later - which can be easily ignored.

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On non-negative weighted graphs, the behavior of Modified Dijkstra's implementation is exactly the same as the Original Dijkstra's so we can use the same time complexity analysis of O((V+E) log V).


However, if the graph has at least one negative weight edge, the analysis is harder.

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当输入图包含至少一个负权重边但没有负权重周期时 - 修改后的Dijkstra算法产生正确的答案。

尝试使用其中一个示例图表ModifiedDijkstra(0):CP3 4.18导致Dijkstra(0)出现问题。

在ModifiedDijkstra算法执行结束时,顶点4具有正确的D [4]值,因为虽然修改后的Dijkstra算法也开始“错误地”认为子路径0→1→3是权重1 + 2 = 3的更好的子路径,因此在调用relax(3,4,3)后使D [4] = 6。这里,修改后的Dijkstra算法在发现其他子路径0→2→3最终是权重10-10 = 0的更好子路径之后继续传播D [3] = 0.因此D [4]最终再次正确。然而,这是以比O((V + E)log V)可能运行(更多)操作为代价的。
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Unfortunately, running ModifiedDijkstra(0) on the graph with negative weight cycle as shown on one of the Example Graphs: CP3 4.17 above will cause an endless loop (the animation is very long but we limit the number of loop to be 100 edges processed so your web browser will not hang).

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Try ModifiedDijkstra(0) on the extreme corner case above that is very hard to derive without proper understanding of this algorithm and was part of Asia Pacific Informatics Olympiad (APIO) 2013 task set by Steven.


The ModifiedDijkstra's algorithm will terminate with correct answer, but only after running exponential number of operations (each carefully constructed triangle raises the number of required operations by another power of two). Thus we cannot prematurely terminate ModifiedDijkstra's in this worst case input situation.


However, such extreme corner case is rare and thus in practice, Modified Dijkstra's algorithm can be used on directed graphs that have some negative weighted edges as long as the graph has no negative weight cycle reachable from the source vertex s.

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The O(V) Depth-First Search (DFS) algorithm can solve special case of SSSP problem, i.e. when the input graph is a (weighted) Tree.


In a Tree, there is only one unique and acylic path that connects two distinct vertices. Thus the unique path that connects the source vertex s to any another vertex uV is actually also the shortest path. For example, try DFS(0) on the Tree above.


Notice that for a (weighted) Tree, we can also use BFS. For example, try BFS(0) on the same Tree above.


Discussion: Why DFS (and also BFS) runs in O(V) instead of O(V+E) if the input is a (weighted) Tree?

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在不是Tree的任何其他图形上运行时,DFS很可能会产生错误的答案。我们将为此类案例显示警告消息,但我们不会阻止您出于教学目的而尝试此功能。

例如,在上面的一般图表上尝试DFS(0),您将看到顶点{4}将具有错误的D [4]值(以及错误的p [4]值),因为DFS(0)深入0→1→首先是3→4,一直回溯到顶点0并且最终访问0→2但是由于DFS之前已经访问过顶点4,所以不能处理边缘2→4。

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The O(V+E) Dynamic Programming algorithm can solve special case of SSSP problem, i.e. when the input graph is a Directed Acyclic Graph (DAG) thus we can find at least one topological order of the DAG and process the edge relaxation according to this topological order.


For example, try DP(0) on the example DAG above. First, it computes one (there are other) possible topological order using either the O(V+E) DFS or the BFS/Kahn's algorithm outlined in Graph Traversal module. For example, assume one topological order is {0,2,1,3,4,5}. Then, it relaxes the outgoing edges of vertices listed in that topological order. After just one O(V+E) pass, we will have correct D[u] values ∀uV.

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On the Modified Dijkstra's killer example shown above, DP(0) works fast as the graph is actually a DAG, albeit having negative weight edge. As the graph is a DAG, there will not be any negative weight cycle to worry about.


However, DP will not work for any non DAG as non DAG contains at least one cycle and thus no topological order can be found within that cycle.

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DP algorithm for solving SSSP on DAG is also called one-pass Bellman Ford's algorithm as it replaces the outermost V-1 loop (we do not know the correct order so we just repeat until the maximum possible) with just one topological order pass (we know that this is (one of) the correct order(s) of this DAG).


Compare DP(0) (relax E edges just once — according to topological order of its vertices) versus BellmanFord(0) (relax E edges in random order, V-1 times) on the same example DAG above.

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We have lots of other stuffs on top of this basic explanation of SSSP algorithms for SSSP problems.


To get BellmanFordDemo.java/cpp, DijkstraDemo.java/cpp, visit Competitive Programming textbook website and get ch4.zip.

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有关此SSSP问题及其各种算法的一些有趣问题,请在 SSSP 培训模块中练习(无需登陆)。


但是,对于注册用户,您应该登陆然后转到 Main Training Page 以正式清除此模块(清除其它先决模块),此类成就将记录在你的账户中。

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We also have a few programming problems that somewhat requires the usage of the correct SSSP algorithm: Kattis - hidingplaces and Kattis - shortestpath1.


Try to solve them and then try the many more interesting twists/variants of this interesting SSSP problem.


Advertisement: Buy Competitive Programming textbook to read more on this interesting problem.

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当操作进行时,状态面板将会有每个步骤的描述。
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e-Lecture: The content of this slide is hidden and only available for legitimate CS lecturer worldwide. Drop an email to visualgo.info at gmail dot com if you want to activate this CS lecturer-only feature and you are really a CS lecturer (show your University staff profile).

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使用用户控件控制动画!可用的快捷键有:
空格键:绘制/停止/重绘
左/右箭头:上一步/下一步
-/+:减缓/增加速度

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Return to 'Exploration Mode' to start exploring!


Note that if you notice any bug in this visualization or if you want to request for a new visualization feature, do not hesitate to drop an email to the project leader: Dr Steven Halim via his email address: stevenhalim at gmail dot com.

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图示

贝尔曼福特的(s)

BFS 算法(s)

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CP3 4.3 U/U

CP3 4.4 D/U

CP3 4.17 D/W

CP3 4.18 -ve weight

CP3 4.19 -ve cycle

CP3 4.40 Tree

Bellman Ford's Killer

Dijkstra's Killer

DAG

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关于 团队 使用条款

关于

VisuAlgo在2011年由Steven Halim博士概念化,作为一个工具,帮助他的学生更好地理解数据结构和算法,让他们自己和自己的步伐学习基础。
VisuAlgo包含许多高级算法,这些算法在Steven Halim博士的书(“竞争规划”,与他的兄弟Felix Halim博士合作)和其他书中讨论。今天,一些高级算法的可视化/动画只能在VisuAlgo中找到。
虽然专门为新加坡国立大学(NUS)学生采取各种数据结构和算法类(例如CS1010,CS1020,CS2010,CS2020,CS3230和CS3230),作为在线学习的倡导者,我们希望世界各地的好奇心发现这些可视化也很有用。
VisuAlgo不是从一开始就设计为在小触摸屏(例如智能手机)上工作良好,因为需要满足许多复杂的算法可视化,需要大量的像素和点击并拖动手势进行交互。一个令人尊敬的用户体验的最低屏幕分辨率为1024x768,并且只有着陆页相对适合移动设备。
VisuAlgo是一个正在进行的项目,更复杂的可视化仍在开发中。
最令人兴奋的发展是自动问题生成器和验证器(在线测验系统),允许学生测试他们的基本数据结构和算法的知识。这些问题是通过一些规则随机生成的,学生的答案会在提交给我们的评分服务器后立即自动分级。这个在线测验系统,当它被更多的世界各地的CS教师采用,应该技术上消除许多大学的典型计算机科学考试手动基本数据结构和算法问题。通过在通过在线测验时设置小(但非零)的重量,CS教练可以(显着地)增加他/她的学生掌握这些基本问题,因为学生具有几乎无限数量的可以立即被验证的训练问题他们参加在线测验。培训模式目前包含12个可视化模块的问题。我们将很快添加剩余的8个可视化模块,以便VisuAlgo中的每个可视化模块都有在线测验组件。
另一个积极的发展分支是VisuAlgo的国际化子项目。我们要为VisuAlgo系统中出现的所有英语文本准备一个CS术语的数据库。这是一个很大的任务,需要众包。一旦系统准备就绪,我们将邀请VisuAlgo游客贡献,特别是如果你不是英语母语者。目前,我们还以各种语言写了有关VisuAlgo的公共注释:
zh, id, kr, vn, th.

团队

项目领导和顾问(2011年7月至今)
Dr Steven Halim, Senior Lecturer, School of Computing (SoC), National University of Singapore (NUS)
Dr Felix Halim, Software Engineer, Google (Mountain View)

本科生研究人员 1 (Jul 2011-Apr 2012)
Koh Zi Chun, Victor Loh Bo Huai

最后一年项目/ UROP学生 1 (Jul 2012-Dec 2013)
Phan Thi Quynh Trang, Peter Phandi, Albert Millardo Tjindradinata, Nguyen Hoang Duy

最后一年项目/ UROP学生 2 (Jun 2013-Apr 2014)
Rose Marie Tan Zhao Yun, Ivan Reinaldo

本科生研究人员 2 (May 2014-Jul 2014)
Jonathan Irvin Gunawan, Nathan Azaria, Ian Leow Tze Wei, Nguyen Viet Dung, Nguyen Khac Tung, Steven Kester Yuwono, Cao Shengze, Mohan Jishnu

最后一年项目/ UROP学生 3 (Jun 2014-Apr 2015)
Erin Teo Yi Ling, Wang Zi

最后一年项目/ UROP学生 4 (Jun 2016-Dec 2017)
Truong Ngoc Khanh, John Kevin Tjahjadi, Gabriella Michelle, Muhammad Rais Fathin Mudzakir

List of translators who have contributed ≥100 translations can be found at statistics page.

致谢
这个项目是由来自NUS教学与学习发展中心(CDTL)的慷慨的教学增进赠款提供的。

使用条款

VisuAlgo是地球上的计算机科学社区免费。如果你喜欢VisuAlgo,我们对你的唯一的要求就是通过你知道的方式,比如:Facebook、Twitter、课程网页、博客评论、电子邮件等告诉其他计算机方面的学生/教师:VisuAlgo网站的神奇存在
如果您是数据结构和算法学生/教师,您可以直接将此网站用于您的课程。如果你从这个网站拍摄截图(视频),你可以使用屏幕截图(视频)在其他地方,只要你引用本网站的网址(http://visualgo.net)或出现在下面的出版物列表中作为参考。但是,您不能下载VisuAlgo(客户端)文件并将其托管在您自己的网站上,因为它是剽窃。到目前为止,我们不允许其他人分叉这个项目并创建VisuAlgo的变体。使用(客户端)的VisuAlgo的离线副本作为您的个人使用是很允许的。
请注意,VisuAlgo的在线测验组件本质上具有沉重的服务器端组件,并且没有简单的方法来在本地保存服务器端脚本和数据库。目前,公众只能使用“培训模式”来访问这些在线测验系统。目前,“测试模式”是一个更受控制的环境,用于使用这些随机生成的问题和自动验证在NUS的实际检查。其他感兴趣的CS教练应该联系史蒂文如果你想尝试这样的“测试模式”。
出版物名单
这项工作在2012年ACM ICPC世界总决赛(波兰,华沙)和IOI 2012年IOI大会(意大利Sirmione-Montichiari)的CLI讲习班上进行了简要介绍。您可以点击此链接阅读我们2012年关于这个系统的文章(它在2012年还没有被称为VisuAlgo)。
这项工作主要由我过去的学生完成。最近的最后报告是:Erin,Wang Zi,Rose,Ivan。
错误申报或请求添加新功能
VisuAlgo不是一个完成的项目。 Steven Halim博士仍在积极改进VisuAlgo。如果您在使用VisuAlgo并在我们的可视化页面/在线测验工具中发现错误,或者如果您想要求添加新功能,请联系Dr Steven Halim博士。他的联系邮箱是他的名字加谷歌邮箱后缀:StevenHalim@gmail.com。