SLIDE 1
Introduction
BFS for Single Source Shortest Path
- Dr. Mattox Beckman
Introduction
BFS for Single Source Shortest Path Dr. Mattox Beckman University - - PowerPoint PPT Presentation
BFS for Single Source Shortest Path Dr. Mattox Beckman University - - PowerPoint PPT Presentation
Introduction BFS for Single Source Shortest Path Dr. Mattox Beckman University of Illinois at Urbana-Champaign Department of Computer Science Introduction Objectives Your Objectives: Implement SSSP using BFS Introduction The Algorithm
SLIDE 2
SLIDE 3
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist Parent a b c d e f g
SLIDE 4
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist ∞ ∞ ∞ ∞ ∞ ∞ Parent
- a
b c d e f g
SLIDE 5
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist 1 1 1 ∞ ∞ ∞ Parent
- a
a a
- a
b c d e f g
SLIDE 6
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist 1 1 1 2 ∞ ∞ Parent
- a
a a b
- a
b c d e f g
SLIDE 7
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist 1 1 1 2 2 ∞ Parent
- a
a a b c
- a
b c d e f g
SLIDE 8
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist 1 1 1 2 2 ∞ Parent
- a
a a b c
- a
b c d e f g
SLIDE 9
Introduction
The Algorithm
◮ Use this if your graph is unweighted ◮ Create a distance array and a parent array a b c d e f g Dist 1 1 1 2 2 3 Parent
- a
a a b c e a b c d e f g
SLIDE 10
Introduction
Implementation
0 // Credit: Competitive Programming 3 1 vi dist(V, INF); dist[s] = 0; 2 queue<int> q; q.push(s); 3 vi parent; 4 while (!q.empty()) { 5
int u = q.front(); q.pop();
6
for (int j = 0; j < (int)AdjList[u].size(); j++) {
7
ii v = AdjList[u][j];
8
if (dist[v.first] == INF) {
9
dist[v.first] = dist[u] + 1;
10
parent[v.first] = u;
11
q.push(v.first);
12 } } }