Shortest Paths Section 4.44.5 Dr. Mayfield and Dr. Lam Department - - PowerPoint PPT Presentation

Shortest Paths

Section 4.4–4.5

• Dr. Mayfield and Dr. Lam

Department of Computer Science James Madison University

Nov 6, 2015

For SSSP on Weighted Graph but without Negative Weight Cycle

DIJKSTRA’s

For SSSP on Weighted Graph but without Negative Weight Cycle

CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS Nov 6, 2015 Shortest Paths 2 of 15

Single Source Shortest Paths (1) Single‐Source Shortest Paths (1)

• If the graph is un weighted, we can use BFS

– But what if the graph is weighted? g p g

• UVa 341 (Non Stop Travel)

(( ) )

• Solution: Dijkstra O((V+E) log V)

– A Greedy Algorithm y g – Use Priority Queue

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Modified Dijkstra’s Example (1) Modified Dijkstra s – Example (1)

3

3

pq = {(0, 2)}

2

2

1 3

5 7

• We store this pair of information to the

priority queue: (D[vertex], vertex), sorted by increasing D[vertex], and

6 5 6 1

• then if ties, by vertex number

See that our priority queue is “clean” at the beginning of (modified) Dijsktra’s

4

1

• the beginning of (modified) Dijsktra s

algorithm, it only contains (0, the source s)

4

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Modified Dijkstra’s Example (2) Modified Dijkstra s – Example (2)

3

3

pq = {(0, 2)} {(2 1) (6 0) (7 3)}

2 1 3

5 7 2

7 2 pq = {(2, 1), (6, 0), (7, 3)}

We greedily take the vertex in the front of the queue (here, it is vertex 2, the source),

6 5 6 1

6

e queue ( e e, s e e , e sou ce), and then successfully relax all its neighbors (vertex 0, 1, 3). P i it Q ill d th 3 ti

4

1

• Priority Queue will order these 3 vertices as

1, 0, 3, with shortest path estimate of 2, 6, 7, respectively.

4

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Modified Dijkstra’s Example (3) Modified Dijkstra s – Example (3)

Vertex 3 appears twice in the priority queue, but this does

pq = {(0, 2)} {(2 1) (6 0) (7 3)}

3

3

not matter, as we will take

• nly the first (smaller) one

pq = {(2, 1), (6, 0), (7, 3)} pq = {(5, 3), (6, 0), (7, 3), (8, 4)}

2 1 3

5 7 2

5 2

We greedily take the vertex in the front of

6 5 6 1

6

We greedily take the vertex in the front of the queue (now, it is vertex 1), then successfully relax all its neighbors (vertex 3 and 4).

4

1

8

) Priority Queue will order the items as 3, 0, 3, 4 with shortest path estimate of 5 6 7 8 respectively

4

8

5, 6, 7, 8, respectively.

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Modified Dijkstra’s Example (4) Modified Dijkstra s – Example (4)

pq = {(0, 2)} {(2 1) (6 0) (7 3)}

3

3

pq = {(2, 1), (6, 0), (7, 3)} pq = {(5, 3), (6, 0), (7, 3), (8, 4)} pq = {(6, 0), (7, 3), (8, 4)}

2 1 3

5 7 2

5 2

6 5 6 1

6

We greedily take the vertex in the front of the queue (now, it is vertex 3), then try to relax all its neighbors (only vertex 4).

4

1

8

relax all its neighbors (only vertex 4). However D[4] is already 8. Since D[3] + w(3, 4) = 5 + 5 is worse than 8, we do not do anything.

4

8

Priority Queue will now have these items 0, 3, 4 with shortest path estimate of 6, 7, 8, respectively. , , , p y

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Modified Dijkstra’s Example (5) Modified Dijkstra s – Example (5)

pq = {(0, 2)} {(2 1) (6 0) (7 3)}

3

3

pq = {(2, 1), (6, 0), (7, 3)} pq = {(5, 3), (6, 0), (7, 3), (8, 4)} pq = {(6, 0), (7, 3), (8, 4)}

2 1 3

5 7 2

5 2 pq = {(7, 3), (7, 4), (8, 4)}

6 5 6 1

6

We greedily take the vertex in the front of the queue (now it is vertex 5) then

4

1

7

the queue (now, it is vertex 5), then successfully relax all its neighbors (only vertex 4).

4

7

Priority Queue will now have these items 3, 4, 4 with shortest path estimate of 7, 7, 8, respectively.

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Modified Dijkstra’s Example (6) Modified Dijkstra s – Example (6)

Remember that vertex 3 appeared twice in the priority queue but this Dijkstra’s

pq = {(0, 2)} {(2 1) (6 0) (7 3)}

3

3

queue, but this Dijkstra s algorithm will only consider the first (shorter) one

pq = {(2, 1), (6, 0), (7, 3)} pq = {(5, 3), (6, 0), (7, 3), (8, 4)} pq = {(6, 0), (7, 3), (8, 4)}

2 1 3

5 7 2

5 2 pq = {(7, 3), (7, 4), (8, 4)} pq = {(7, 4), (8, 4)} pq = {(8, 4)}

6 5 6 1

6 pq {( , )} pq = {}

4

1

7

Similarly for vertex 4. The one with shortest path estimate 7

4

7

will be processed first and the

• ne with shortest path

estimate 8 will be ignored, although nothing is changed although nothing is changed anymore

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Dijkstra’s Algorithm (using STL) Dijkstra s Algorithm (using STL)

i di ( ) di [ ] // 2 vi dist(V, INF); dist[s] = 0; // INF = 2B priority_queue< ii, vector<ii>, greater<ii> > pq; pq.push(ii(0, s)); // sort based on increasing distance while (!pq.empty()) { // main loop ii top = pq.top(); pq.pop(); // greedy int d = top.first, u = top.second; p , p ; if (d == dist[u]) { for (int j = 0; j < (int)AdjList[u].size(); j++) { ii v = AdjList[u][j]; // all outgoing edges from u ii v = AdjList[u][j]; // all outgoing edges from u if (dist[u] + v.second < dist[v.first]) { dist[v.first] = dist[u] + v.second; // relax ii i i i pq.push(ii(dist[v.first], v.first)); } // enqueue this neighbor regardless it is } // already in pq or not } }

CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS Nov 6, 2015 Shortest Paths 10 of 15

int[] dist = new int[adjList.length]; for (int i = 0; i < dist.length; i++) { dist[i] = 99999999; } dist[src] = 0; PriorityQueue<Edge> pq = new PriorityQueue<Edge>(); pq.add(new Edge(0, src)); while (pq.size() > 0) { Edge top = pq.poll(); int u = top.dest; if (top.weight == dist[u]) { for (int j = 0; j < adjList[u].length; j++) { Edge v = adjList[u][j]; if (dist[u] + v.weight < dist[v.dest]) { dist[v.dest] = dist[u] + v.weight; pq.add(new Edge(dist[v.dest], v.dest)); } } } }

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For All‐Pairs Shortest Paths

FLOYD WARSHALL’s

For All Pairs Shortest Paths

CS3233 ‐ Competitive Programming, Steven Halim, SoC, NUS Nov 6, 2015 Shortest Paths 11 of 15

UVa 11463 – Commandos (1) ( )

Al‐Khawarizmi, Malaysia National Contest 2008

• Given:

– A table that stores the amount of minutes to travel between buildings (there are at most 100 buildings) – 2 special buildings: startB and endB – K soldiers to bomb all the K buildings in this mission – Each of them start at the same time from startB, h b ildi h h b b b d b h choose one building B that has not been bombed by other soldier (bombing time negligible), and then gather in (destroyed) building endB and then gather in (destroyed) building endB.

• What is the minimum time to complete the mission?

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UVa 11463 – Commandos (2) ( )

Al‐Khawarizmi, Malaysia National Contest 2008

• How long do you need to solve this problem?
• Solution:

– The answer is determined by sp from starting building, detonate furthest building, and sp from that furthest building to end building

• max(dist[start][i] + dist[i][end]) for all i V
• How to compute sp for many pairs of vertices?

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UVa 11463 – Commandos (3) ( )

Al‐Khawarizmi, Malaysia National Contest 2008

• This problem is called: All‐Pairs Shortest Paths
• Two options to solve this:

– Call SSSP algorithms multiple times

• Dijkstra O(V * (V+E) * log V), if E = V2 O(V3 log V)
• Bellman Ford O(V * V * E), if E = V2 O(V4)
• Slow to code

l d h ll l l i h – Use Floyd Warshall, a clever DP algorithm

• O(V3) algorithm
• Very easy to code!
• Very easy to code!
• In this problem, V is <= 100, so Floyd Warshall is DOABLE!!

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Floyd Warshall Template Floyd Warshall – Template

• O(V3) since we have three nested loops!
• Use adjacency matrix: G[MAX_V][MAX_V];

_ _

– So that weight of edge(i, j) can be accessed in O(1)

for (int k = 0; k < V; k++) for (int k = 0; k < V; k++) for (int i = 0; i < V; i++) for (int j = 0; j < V; j++) j j j G[i][j] = min(G[i][j], G[i][k] + G[k][j]);

See more explanation of this three liner DP – See more explanation of this three‐liner DP algorithm in CP

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