BFS

1. FIND ALL EDGES IN A
GRAPH WHICH BELONG
TO ANY SHORTEST PATH
FROM SOURCE TO
Concepts used – BFS only

2. QUESTION
1
2 3 4
56 7
8 9 10 11
12 13
Source Destination
1 56 7
12 13
Source Destination
Graph containing only those
edges belonging to shortest
path from 1 to 5
From To
5 7
5 13
7 6
13 12
6 1
12 1
Solution Edges
List
Expected Time Complexity –
O(V+E)

3. APPROACH
1
2 3 4
56
8 9 10 11
12 13
7
Source Destination
Shortest Path are from 1 to 5
are –
1 – 6 – 7 – 5
1 – 12 – 13 – 5

4. APPROACH
1
2 3 4
56
8 9 10 11
12 13
7
Source Destination
Steps –
1. Apply BFS from Source node.
2. Store edges which are part of
shortest distance to that node.

5. APPROACH
1
2
6
8
12
Source
Child Paren
t
2 1
6 1
8 1
12 1
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1

6. APPROACH
1
2 3
6 7
8 9
12 13
Source
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12

7. APPROACH
1
2 3 4
6 7
8 9 10
12 13
Source
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
10 3
4 3
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
4 3
10 9

8. APPROACH
1
2 3 4
56 7
8 9 10
12 13
Source Destination
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
10 3
4 3
5 3
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7

9. APPROACH
1
2 3 4
56 7
8 9 10
12 13
Source Destination
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
10 3
4 3
5 3
This edge also gives
Shortest distance to 5 =
3 using path 1 – 12 – 13
– 5
So include this edge too
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

10. APPROACH
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
10 3
4 3
5 3
1
2 3 4
56 7
8 9 10 11
12 13
Source Destination
Inclusion of edge from
4 to 5 gives distance
from 1 to 5 = 4 which is
larger than shortest
distance to 5 (equal to
3). So we will not
include it in our parent-
child table.
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

11. APPROACH
Node Shortest
Distance
From
Source
1 0
2 1
6 1
8 1
12 1
3 2
7 2
9 2
13 2
10 3
4 3
5 3
1
2 3 4
56 7
8 9 10 11
12 13
Source Destination
Inclusion of edge from
11 to 5 gives distance
from 1 to 5 = 5 which is
larger than shortest
distance to 5 (equal to
3). So we will not
include it in our parent-
child table.
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

12. APPROACH
1
2 3 4
56 7
8 9 10 11
12 13
Source Destination
Next Step –
Apply BFS/DFS from Destination using
parent child table and store all the
edges which come while traversing the
graph.
Do not continue the BFS/DFS to the
edges of source node. So stop at
source node.
Stop the BFS at when is
Destination. Do not
continue the BFS as it
will include unnecessary
edges.
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

13. APPROACH
57
13
Destination
From To
5 7
5 13
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

14. APPROACH
56 7
12 13
Destination
From To
5 7
5 13
7 6
13 12
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

15. APPROACH
1 56 7
12 13
Source Destination
From To
5 7
5 13
7 6
13 12
6 1
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

16. APPROACH
1 56 7
12 13
Source Destination
Do not explore to
the edges of
Source Node.
From To
5 7
5 13
7 6
13 12
6 1
12 1
Solution Edges
List
Child Paren
t
2 1
6 1
8 1
12 1
3 2
7 6
9 8
13 12
5 7, 13

17. CODE FOR BFS
void bfs1(int source,int destintion)
{
for(int i=0;i<100000;i++)
dis[i]=inf
dis[source]=0;
queue<pair<int,int> > q;// first number is node, second number is distance
q.push(mp(source,0));
while(!q.empty())
{
int node = q.front().first;
int d = q.front().second;
q.pop();
if(node==destination)
break;
for(int i=0;i<v[node].size();i++)
{
int neighbour = v[node][i];
if(d+1<dis[neighbour])
{
dis[neighbour] = d+1;
q.push(mp(neighbour,d+1));
par[neighbour].pb(node)
}
else if(d+1==dis[neighbour])
{
par[neighbour].pb(node);
}
}
}
Global Variables Declared -
#define inf INT_MAX
#define pb push_back
#define mp make_pair
vector<int> v[100000];
vector<int> par[100000];
int dis[100000];
vector<pair<int,int> > sol_edges;

18. void bfs2(int source,int destination)
{
int vis[100000]={0};
queue<int> q;
q.push(destination);
while(!q.empty())
{
int node = q.front();
q.pop();
if(vis[node]!=0)
continue;
if(node==source)
break;
for(int i=0;i<par[node].size();i++)
{
int neighbour = par[node][i];
if(vis[neighbour]!=0)
continue;
sol_edges.pb(mp(node,neighbour));
q.push(neighbour);
}
}
return;
Global Variables Declared -
#define inf INT_MAX
#define pb push_back
#define mp make_pair
vector<int> v[100000];
vector<int> par[100000];
int dis[100000];
vector<pair<int,int> >
sol_edges;
CODE FOR BFS - 2