Maximum Flow Problem
flow
capacity
Source
s
actual
flow
3/ 4
2/ 2
1/ 1
a
b
2/3
d
2/ 2
1/2
c
e
1. Actual flow  capacity
2. The source node can
produce flow as much as
possible
3. For every node other
than s and t,
flow out  flow in
The target node’s flow-in is
called the flow of the graph
f
1/ 2
Target
(sink)
7/13/2017
A node may have different
ways to dispatch its flow-out .
g
2/ 3
t
What is the maximum flow of
the graph?
ACM-ICPC
1
What makes the maximum flow problem more
difficult than the shortest path problem?
Source
s
3/ 4
The shortest path of t is based
on the shortest path of g.
2/ 2
1/ 1
a
once a node’s shortest path
from s is determined, we will
never have to revise the
decision
b
2/3
d
2/ 2
1/2
c
e
The maximum flow to t is
not based on the maximum
flows to f and g.
f
g
1/ 2
Target
7/13/2017
2/ 3
t
ACM-ICPC
2
local optimum does not imply global optimum
Source
Source
s
s
4/5
5/5
2/2
0/1
a
b
3/3
1/4
3/3
c
Source
2/3
d
a
0/3
c
s
2/2
5/5
1/1
b
4/4
3/3
0/3
d
a
1/3
c
2/2
1/1
b
3/4
1/3
3/3
d
Target
Target
sc
max flow = 6
7/13/2017
2/2
sd
max flow = 7
ACM-ICPC
Target
5/5
t
st
max flow = 7
3
A naïve approach: adding all possible paths together
Source
Source
s
s
2/3
a
1
2
1
2
c
d
2/ 2
3
Target
a
b
4
2/ 3
t
Source
s
2/ 2
1
2/ 2
d
c
1
b
1/ 4
1
d
c
2/3
Target
t
Residual Graph
7/13/2017
a
b
4
1
1/1
ACM-ICPC
1/1
Target
t
Residual Graph
4
Source
Source
s
s
3/3
a
1
b
a
3
1
d
Target
b
2/2
c
d
2/ 2
3/3
t
Target
t
Not always work.
We are just lucky
Residual Graph
7/13/2017
1
1/4
3
c
2/2
ACM-ICPC
5
We may have a wrong choice:
Source
Source
s
3/3
2
2
1
a
b
3/4
3
2
Target
2
b
1
3
2
d
c
3/3
t
1
a
d
c
7/13/2017
s
2
So, we keep a path for
changing decision:
ACM-ICPC
Target
t
6
Keep a path for changing decision:
Source
Source
s
3/3
s
2
2
3
1
a
b
3/4
3
1
a
2
b
1
3
2
3
d
c
2
Target
d
c
3/3
2
t
Target
3
t
Augmented Graph
7/13/2017
ACM-ICPC
7
Find another path in the augmented graph
No more path
from s to t
Source
s
s
2/2
3
a
b
2/3
2/ 2
3
t
Augmented Graph
7/13/2017
3
b
a
2
1
b
2
1
2
1
2
d
c
2
1
a
2/2
2/3
Target
3
1
1
s
1
2
d
c
2
3
1
3
2
1
d
c
2
t
3
t
new Augmented Graph
ACM-ICPC
8
Add flows together
Source
3
+
s
=
s
3
2
a
b
3
a
b
2
2
d
2
a
b
2
2
c
s
2
1
d
c
d
c
3
3
2
Target
2
t
Target
t
Target
t
Maximum Flow
7/13/2017
ACM-ICPC
9
An algorithm for finding the maximum flow of graphs
Input: G = (V, E), and s,t  V
Let Gmax = (, ), Gaug = G
1. Find a path p from s to t in Gaug
2. If no such p exists, output Gmax and stop the
program
3. Add p into Gmax and update Gaug
4. Repeat 1, 2, and, if possible, 3
Complexity:
7/13/2017
O(f |E|)
ACM-ICPC
10
Claim: This algorithm always terminates
with a maximum flow of the input graph
The proof is somewhat difficult
and beyond the scope of 279.
Input: G = (V, E), and s,t  V
Let Gmax = (, ), Gaug = G
1.
2.
3.
4.
I say: No, it is not difficult and
every one here should know.
Find a path p from s to t in Gaug
If no such p exists, output Gmax and stop the program
Add p into Gmax and update Gaug
Repeat 1, 2, and 3 if possible
•
Since any given graph has a finite maximum flow, and every path
from s to t, if any, provides a positive flow, the program therefore
cannot run forever.
•
By contradiction, suppose the algorithm terminates with a flow in
Gmax. that is not maximum.
Then, there must be a flow not included in Gmax. But, if this is the
case, there must be a path from s to t to carries this flow, and
hence the program should not terminate at this moment. A
contradiction.
7/13/2017
ACM-ICPC
11
An inefficient situation
Source
s
1/1000
1/1000
1/1
a
b
1/ 1
1/1000
1/ 1000
t
Target
7/13/2017
ACM-ICPC
12
An inefficient situation
Source
s
2/ 1/1000
2/ 1/ 1000
1/ 1
a
b
2/ 1/1000
2/ 1/ 1000
t
Target
7/13/2017
ACM-ICPC
13
Solution
Source
s
1000/ 1000
Always chose the
maximum next edge
1000
a
b
1000/ 1000
1/ 1000
t
Target
7/13/2017
ACM-ICPC
14