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[ACM-ICPC] Minimum Cut
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KuoE0
January 21, 2013
Programming
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[ACM-ICPC] Minimum Cut
KuoE0
January 21, 2013
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Transcript
Minimum Cut ֲࢸݢʢKuoE0ʣ
[email protected]
KuoE0.ch
Cut
cut (undirected)
1 3 5 6 7 8 9 2 4 undirected
graph A partition of the vertices of a graph into two disjoint subsets
1 3 5 6 7 8 9 2 4 undirected
graph A partition of the vertices of a graph into two disjoint subsets
1 3 5 6 7 8 9 2 4 undirected
graph A partition of the vertices of a graph into two disjoint subsets
1 2 8 5 4 7 9 3 6 A
partition of the vertices of a graph into two disjoint subsets undirected graph
1 2 8 5 4 7 9 3 6 Cut-set
of the cut is the set of edges whose end points are in different subsets. undirected graph
1 2 8 5 4 7 9 3 6 Cut-set
of the cut is the set of edges whose end points are in different subsets. Cut-set undirected graph
1 2 8 5 4 7 9 3 6 weight
= number of edges or sum of weight on edges weight is 7 undirected graph
cut (directed)
1 3 5 6 7 8 9 2 4 directed
graph A partition of the vertices of a graph into two disjoint subsets
1 3 5 6 7 8 9 2 4 directed
graph A partition of the vertices of a graph into two disjoint subsets
1 3 5 6 7 8 9 2 4 directed
graph A partition of the vertices of a graph into two disjoint subsets
1 2 8 5 4 7 9 3 6 directed
graph A partition of the vertices of a graph into two disjoint subsets
1 2 8 5 4 7 9 3 6 directed
graph Cut-set of the cut is the set of edges whose end points are in different subsets.
1 2 8 5 4 7 9 3 6 directed
graph Cut-set of the cut is the set of edges whose end points are in different subsets.
1 2 8 5 4 7 9 3 6 Cut-set
directed graph Cut-set of the cut is the set of edges whose end points are in different subsets.
1 2 8 5 4 7 9 3 6 weight
is 5⇢ or 2⇠ directed graph weight = number of edges or sum of weight on edges
s-t cut 1. one side is source 2. another side
is sink 3. cut-set only consists of edges going from source’s side to sink’s side
1 3 5 6 7 8 9 2 4 flow
network Source Sink Other
1 3 5 6 7 8 9 2 4 flow
network Source Sink
1 3 5 6 7 8 9 2 4 flow
network Source Sink
1 2 8 5 4 7 9 3 6 flow
network cut-set only consists of edges going from source’s side to sink’s side
1 2 8 5 4 7 9 3 6 weight
is 6 flow network cut-set only consists of edges going from source’s side to sink’s side
Max-Flow Min-Cut Theorem
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 3 + 3 = 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 3 + 3 = 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 3 + 4 - 1 = 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 3 + 4 - 1 = 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 4 + 2= 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 4 + 2= 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 4 + 2= 6
Observation 1 The network flow sent across any cut is
equal to the amount reaching sink. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 total flow = 6, flow on cut = 4 + 2= 6
Observation 2 Then the value of the flow is at
most the capacity of any cut. 1 3 2 4 5 6 3 8 4 2 4 4 3 It’s trivial!
Observation 2 Then the value of the flow is at
most the capacity of any cut. 1 3 2 4 5 6 3 8 4 2 4 4 3 It’s trivial!
Observation 3 Let f be a flow, and let (S,T)
be an s-t cut whose capacity equals the value of f. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 f is the maximum flow (S,T) is the minimum cut
Observation 3 Let f be a flow, and let (S,T)
be an s-t cut whose capacity equals the value of f. 1 3 2 4 5 6 3/3 3/8 4/4 2/2 1/4 4/4 2/3 f is the maximum flow (S,T) is the minimum cut
Max-Flow EQUAL Min-Cut
Example
1 3 2 4 5 6 3 8 4 2
4 4 3
1 3 2 4 5 6 3/3 3/8 4/4 2/2
1/4 4/4 2/3 Maximum Flow = 6
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Residual Network
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Minimum Cut = 6
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Minimum Cut = 6
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Minimum Cut = 6
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Minimum Cut = 6
The minimum capacity limit the maximum flow!
find a s-t cut
1 3 2 4 5 6 3/3 3/8 4/4 2/2
1/4 4/4 2/3 Maximum Flow = 6
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 Travel on Residual Network
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 start from source
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 don’t travel through full edge
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 don’t travel through full edge
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 no residual edge
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 no residual edge
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 s-t cut
1 3 2 4 5 6 0/3 5/8 0/4 0/2
3/4 0/4 1/3 s-t cut
result of starting from sink 1 3 2 4 5
6 0/3 5/8 0/4 0/2 3/4 0/4 1/3
result of starting from sink 1 3 2 4 5
6 0/3 5/8 0/4 0/2 3/4 0/4 1/3
Minimum cut is non-unique!
time complexity: based on max-flow algorithm Ford-Fulkerson algorithm O(EF) Edmonds-Karp
algorithm O(VE2) Dinic algorithm O(V2E)
Stoer Wagner only for undirected graph time complexity: O(N3) or
O(N2log2N)
UVa 10480 - Sabotage Practice Now
Problem List UVa 10480 UVa 10989 POJ 1815 POJ 2914
POJ 3084 POJ 3308 POJ 3469
Reference • http://www.flickr.com/photos/dgjones/335788038/ • http://www.flickr.com/photos/njsouthall/3181945005/ • http://www.csie.ntnu.edu.tw/~u91029/Cut.html • http://en.wikipedia.org/wiki/Cut_(graph_theory) •
http://en.wikipedia.org/wiki/Max-flow_min-cut_theorem • http://www.cs.princeton.edu/courses/archive/spr04/cos226/lectures/ maxflow.4up.pdf • http://www.cnblogs.com/scau20110726/archive/ 2012/11/27/2791523.html
Thank You for Your Listening.