all pair shortest path ioi/acm icpc training june 2004

All Pair Shortest Path

IOI/ACM ICPC Training

June 2004

All Pair Shortest Path

Note: Dijkstra’s Algorithm takes O((V+E)logV) time

All Pair Shortest Path Problem can be solved by executing Dijkstra’s Algorithm |V| times

Running Time: O(V(V+E)log V)

Floyd-Warshall Algorithm: O(V3)

Idea

Label the vertices with integers 1..n

Restrict the shortest paths from i to j to consist of vertices 1..k only (except i and j)

Iteratively relax k from 1 to n.

i

k

j

Definition

Find shortest distance from i to j using vertices 1 .. k only

i

k

j

Example

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3 5

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1

53

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3

i = 4, j = 5, k = 0

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3 5

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1

53

1

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i = 4, j = 5, k = 1

1

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3 5

2 4

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1

53

1

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i = 4, j = 5, k = 2

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3 5

2 4

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1

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i = 4, j = 5, k = 3

1

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3 5

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51

1

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Idea

k

j

i

0 k for

0 k for ),min(

only {1..k} involving to from distanceShortest :

,

1,

1,

1,

,

,

ji

kjk

kki

kjik

ji

kji

w

DDDD

jiD

The tables

i

j

k=4

k=5

i

j

k

k

The code

for i = 1 to |V| for j = 1 to |V| a[i][j][0] = cost(i,j)for k = 1 to |V| for i = 1 to |V| for j = 1 to |V| a[i][j][k] = min( a[i][j][k-1],

a[i][k][k-1] + a[k][j][k-1])

Topological sort


June 2004

Topological order

Consider the prerequisite structure for courses:

Each node x represents a course x (x, y) represents that course x is a prerequisite to course y Note that this graph should be a directed graph without cycles. A linear order to take all 5 courses while satisfying all prerequisites is called a

topological order. E.g.

a, c, b, e, d c, a, b, e, d

b d

ec

a

Topological sort

Arranging all nodes in the graph in a topological order

Applications:Schedule tasks associated with a project

Topological sort algorithm

Algorithm topSort1n = |V|;Let R[0..n-1] be the result array;for i = 1 to n {

select a node v that has no successor;R[n-i] = v;delete node v and its edges from the graph;

}return R;

Example

b d

ec

a

1. d has no successor! Choose d!

a

5. Choose a!The topological

order is a,b,c,e,d

2. Both b and e have no successor! Choose e!

b

ec

a

3. Both b and c have no successor! Choose c!

b

c

a

4. Only b has no successor! Choose b!

ba

Time analysis

Finding a node with no successor takes O(|V|+|E|) time.

We need to repeat this process |V| times. Total time = O(|V|2 + |V| |E|).

We can implement the above process using DFS. The time can be improved to O(|V| + |E|).

Algorithm based on DFSAlgorithm topSort2s.createStack();for (all nodes v in the graph) {

if (v has no predecessors) {s.push(v);mark v as visited;

}}while (s is not empty) {

let v be the node on the top of the stack s;if (no unvisited nodes are children to v) { // i.e. v has no unvisited successor

aList.add(1, v);s.pop(); // blacktrack

} else {select an unvisited child u of v;s.push(u);mark u as visited;

}}return aList;

Bipartite Matching


June 2004

Unweighted Bipartite Matching

Definitions

Matching

Free Vertex

Definitions

Maximum Matching: matching with the largest number of edges

Definition

Note that maximum matching is not unique.

Intuition

Let the top set of vertices be men Let the bottom set of vertices be women Suppose each edge represents a pair of

man and woman who like each other

Maximum matching tries to maximize the number of couples!

Applications

Matching has many applications. For examples,Comparing Evolutionary TreesFinding RNA structure…

This lecture lets you know how to find maximum matching.

Alternating Path Alternating between matching and non-matching edges.

a b c d e

f g h i j

d-h-e: alternating patha-f-b-h-d-i: alternating path starts and ends with free verticesf-b-h-e: not alternating pathe-j: alternating path starts and ends with free vertices

Idea

“Flip” augmenting path to get better matching

Note: After flipping, the number of matched edges will increase by 1!

Idea

Theorem (Berge 1975):A matching M in G is maximum iffThere is no augmenting path

Proof: () If there is an augmenting path, clearly not

maximum. (Flip matching and non-matching edges in that path to get a “better” matching!)

Proof for the other direction

() Suppose M is not maximum. Let M’ be a maximum matching such that |M’|>|M|.

Consider H = MM’ = (MM’)-(MM’)i.e. a set of edges in M or M’ but not both

H has two properties: Within H, number of edges belong to M’ > number of edges

belong to M. H can be decomposed into a set of paths. All paths should

be alternating between edges in M and M’. There should exist a path with more edges from M’.

Also, it is alternating.

Idea of Algorithm

Start with an arbitrary matching While we still can find an augmenting path

Find the augmenting path PFlip the edges in P

Labelling Algorithm

Start with arbitrary matching

Labelling Algorithm

Pick a free vertex in the bottom

Labelling Algorithm

Run BFS

Labelling Algorithm

Alternate unmatched/matched edges

Labelling Algorithm

Until a augmenting path is found

Augmenting Tree

Repeat

Pick another free vertex in the bottom

Repeat

Run BFS

Repeat

Flip

Answer

Since we cannot find any augmenting path, stop!

Overall algorithm

Start with an arbitrary matching (e.g., empty matching) Repeat forever

For all free vertices in the bottom, do bfs to find augmenting paths

If found, then flip the edges If fail to find, stop and report the maximum matching.

Time analysis

We can find at most |V| augmenting paths (why?)

To find an augmenting path, we use bfs! Time required = O( |V| + |E| )

Total time: O(|V|2 + |V| |E|)

Improvement

We can try to find augmenting paths in parallel for all free nodes in every iteration.

Using such approach, the time complexity is improved to O(|V|0.5 |E|)

Weighted Bipartite Graph

46 63

Weighted Matching

46 6

3

Score: 6+3+1=10

Maximum Weighted Matching

46 6

3

Score: 6+1+1+1+4=13

Augmenting Path (change of definition) Any alternating path such that total score of unmatched

edges > that of matched edges The score of the augmenting path is

Score of unmatched edges – that of matched edges

46 6

3

Note: augmenting path need not start and end at free vertices!

Idea for finding maximum weight matching Theorem: Let M be a matching of

maximum weight among matchings of size |M|.

If P is an augmenting path for M of maximum weight,

Then, the matching formed by augmenting M by P is a matching of maximum weight among matchings of size |M|+1.

Overall Algorithm

Start with an empty matching Repeat forever

Find an augmenting path P with maximum score If the score > 0, then flip the edges Otherwise, stop and report the maximum weight

matching.

Time analysis

The same! Time required = O(|V|2 + |V| |E|)

Stable Marriage Problem


June 2004

Stable Marriage Problem

Given N men and N women, each person list in order of preference all the people of the opposite sex who would like to marry.

Problem:Engage all the women to all the men in such a

way as to respect all their preferences as much as possible.

Stable?

A set of marriages is unstable if two people who are not married both prefer each other than their

spouses

E.g. Suppose we have A1 B3 C2 D4 E5. This is unstable since A prefer 2 more than 1 2 prefer A more than C

A B C D E

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A

Naïve solution

Starting from a feasible solution. Check if it is stable.

If yes, done! If not, remove an unstable couple.

Is this work?

Naïve solution (2)

Does not work! E.g.

A1 B3 C2 D4 E5 A2 B3 C1 D4 E5 A3 B2 C1 D4 E5 A3 B1 C2 D4 E5

A B C D E

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A

Solution

1. Let X be the first man.2. X proposes to the best woman in the

remaining on his list. (Initially, the first woman on his list!)

3. If α is not engaged Pair up (X, α). Then, set X=next man and goto 1.

4. If α prefers X more than her fiancee Y, Pair up (X, α). Then, set X=Y and goto 1.

5. Goto 1

Example

A B C D E

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A B C D E

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Time analysis

If there are N men and N women,O(N2) time

Algorithm

prefer[m][s]=w means the woman w is on the s-th position in the preference list of the man m

Let next[m] be the current best woman in his remaining list. (Initially, next[m]=0)

fiancee[w]=m means the man m engaged to woman w. (Initially, fiancee[w]=0)

Let rank[w][m] is the ranking of the man m in the preference list of the woman w.

For(m=1;m<=N;m++) {For(s=m;s!=0;

}

all pair shortest path ioi/acm icpc training june 2004

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