Need help on a problemset in a programming contest [closed] - algorithm

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I've attended a local programming contest on my country. The name of the contest is "ACM-ICPC Indonesia National Contest 2013".
The contest has ended on 2013-10-13 15:00:00 (GMT +7) and I am still curious about one of the problems.
You can find the original version of the problem here.
Brief Problem Explanation:
There are a set of "jobs" (tasks) that should be performed on several "servers" (computers).
Each job should be executed strictly from start time Si to end time Ei
Each server can only perform one task at a time.
(The complicated thing goes here) It takes some time for a server to switch from one job to another.
If a server finishes job Jx, then to start job Jy it will need an intermission time Tx,y after job Jx completes. This is the time required by the server to clean up job Jx and load job Jy.
In other word, job Jy can be run after job Jx if and only if Ex + Tx,y ≤ Sy.
The problem is to compute the minimum number of servers needed to do all jobs.
Example:
For example, let there be 3 jobs
S(1) = 3 and E(1) = 6
S(2) = 10 and E(2) = 15
S(3) = 16 and E(3) = 20
T(1,2) = 2, T(1,3) = 5
T(2,1) = 0, T(2,3) = 3
T(3,1) = 0, T(3,2) = 0
In this example, we need 2 servers:
Server 1: J(1), J(2)
Server 2: J(3)
Sample Input:
Short explanation: The first 3 is the number of test cases, following by number of jobs (the second 3 means that there are 3 jobs for case 1), then followed by Ei and Si, then the T matrix (sized equal with number of jobs).
3
3
3 6
10 15
16 20
0 2 5
0 0 3
0 0 0
4
8 10
4 7
12 15
1 4
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
4
8 10
4 7
12 15
1 4
0 50 50 50
50 0 50 50
50 50 0 50
50 50 50 0
Sample Output:
Case #1: 2
Case #2: 1
Case #3: 4
Personal Comments:
The time required can be represented as a graph matrix, so I'm supposing this as a directed acyclic graph problem.
Methods I tried so far is brute force and greedy, but got Wrong Answer. (Unfortunately I don't have my code anymore)
Could probably solved by dynamic programming too, but I'm not sure.
I really have no clear idea on how to solve this problem.
So a simple hint or insight will be very helpful to me.

You can solve this by computing the maximum matching in a bipartite graph.
The idea is you are trying to match job end times with job start times.
A matched end time x with start time y means that the same server will do job x and job y.
The number of servers you need will correspond to the number of unmatched start times (because each of these jobs will require a new server).
Example Python code using NetworkX:
import networkx as nx
G=nx.DiGraph()
S=[3,10,16] # start times
E=[6,15,20] # end times
T = [ [0, 2, 5],
[0, 0, 3],
[0, 0, 0] ] # T[x][y]
N=len(S)
for jobx in range(N):
G.add_edge('start','end'+str(jobx),capacity=1)
G.add_edge('start'+str(jobx),'end',capacity=1)
for joby in range(N):
if E[jobx]+T[jobx][joby] <= S[joby]:
G.add_edge('end'+str(jobx),'start'+str(joby),capacity=1)
print N-nx.max_flow(G,'start','end')

Related

Algorithm strategy to prevent values from bouncing between 2 values when on edge of two 'buckets'

I'm tracking various colored balls in OpenCV (Python) in real-time. The tracking is very stable. i.e. when stationary the values do not change with more then 1 / 2 pixels for the center of the circle.
However i'm running into what must surely be a well researched issue: I need to now place the positions of the balls into an rougher grid - essentially simply dividing (+ rounding) the x,y positions.
e.g.
input range is 0 -> 9
target range is 0 -> 1 (two buckets)
so i do: floor(input / 5)
input: [0 1 2 3 4 5 6 7 8 9]
output: [0 0 0 0 0 1 1 1 1 1]
This is fine, but the problem occurs when just a small change in the initial value might result it to be either in quickly changes output single I.e. at the 'edge' of the divisions -or a 'sensitive' area.
input: [4 5 4 5 4 5 5 4 ...]
output:[0 1 0 1 0 1 1 0 ...]
i.e. values 4 and 5 (which falls withing the 1 pixel error/'noisy' margin) cause rapid changes in output.
What are some of the strategems / algorithms that deal with these so help me further?
I searched but it seems i do not how to express the issue correctly for Google (or StackOverflow).
I tried adding 'deadzones'. i.e. rather then purely dividing i leave 'gaps' in my ouput range which means a value sometimes has no output (i.e. between 'buckets'). This somewhat works but means i have a lot (i.e. the range of the fluctuation) of the screen that is not used...
i.e.
input = [0 1 2 3 4 5 6 7 8 9]
output = [0 0 0 0 x x 1 1 1 1]
Temporal averaging is not ideal (and doesn't work too well either) - and increases the latency.
I just have a 'hunch' there is a whole set of Computer / Signal science about this.

Palindrome partitioning with interval scheduling

So I was looking at the various algorithms of solving Palindrome partitioning problem.
Like for a string "banana" minimum no of cuts so that each sub-string is a palindrome is 1 i.e. "b|anana"
Now I tried solving this problem using interval scheduling like:
Input: banana
Transformed string: # b # a # n # a # n # a #
P[] = lengths of palindromes considering each character as center of palindrome.
I[] = intervals
String: # b # a # n # a # n # a #
P[i]: 0 1 0 1 0 3 0 5 0 3 0 1 0
I[i]: 0 1 2 3 4 5 6 7 8 9 10 11 12
Example: Palindrome considering 'a' (index 7) as center is 5 "anana"
Now constructing intervals for each character based on P[i]:
b = (0,2)
a = (2,4)
n = (2,8)
a = (2,12)
n = (6,12)
a = (10,12)
So, now if I have to schedule these many intervals on time 0 to 12 such that minimum no of intervals should be scheduled and no time slot remain empty, I would choose (0,2) and (2,12) intervals and hence the answer for the solution would be 1 as I have broken down the given string in two palindromes.
Another test case:
String: # E # A # B # A # E # A # B #
P[i]: 0 1 0 1 0 5 0 1 0 5 0 1 0 1 0
I[i]: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Plotting on graph:
Now, the minimum no of intervals that can be scheduled are either:
1(0,2), 2(2,4), 5(4,14) OR
3(0,10), 6(10,12), 7(12,14)
Hence, we have 3 partitions so the no of cuts would be 2 either
E|A|BAEAB
EABAE|A|B
These are just examples. I would like to know if this algorithm will work for all cases or there are some cases where it would definitely fail.
Please help me achieve a proof that it will work in every scenario.
Note: Please don't discourage me if this post makes no sense as i have put enough time and effort on this problem, just state a reason or provide some link from where I can move forward with this solution. Thank you.
As long as you can get a partition of the string, your algorith will work.
Recall to mind that a partion P of a set S is a set of non empty subset A1, ..., An:
The union of every set A1, ... An gives the set S
The intersection between any Ai, Aj (with i != j) is empty
Even if the palindrome partitioning deals with strings (which are a bit different from sets), the properties of a partition are still true.
Hence, if you have a partition, you consequently have a set of time intervals without "holes" to schedule.
Choosing the partition with the minimum number of subsets, makes you have the minimum number of time intervals and therefore the minimum number of cuts.
Furthermore, you always have at least one palindrome partition of a string: in the worst case, you get a palindrome partition made of single characters.

Looking for failing test case to DP solution to MARTIAN on SPOJ

I am trying to solve the MARTIAN problem on SPOJ
My algorithm is as follows:
Define dp[i][j]=max amount of minerals that can be mined in the rectangle form 0,0 to i,j.
Use the recurrence
dp[i][j] = max(dp[i-1][j] + total amount of yeyenum
in the i-th row up to the j-th column,
dp[i][j-1] + total amount of bloggium
in the j-th column up to the cell i-th row)
However such an approach yields a WA (Wrong Answer). Can someone please provide me with a test case where such and approach will not work?
I am not looking for the correct algorithm just a test case where this approach fails as. I've been unable to find the bug myself.
Try this on your code(modified from the example given):
4 4
0 0 10 60
1 3 10 0
4 2 1 3
1 1 20 0
10 0 0 0
1 1 1 10
0 0 5 3
5 10 10 10
0 0
If you start by looking at [4][4], you'll choose Bloggium, because you can get 23 bloggium by going up, and only 22 Yeyenum from going left. However, you're going to miss a huge amount of Yeyenum.
Using your algorithm, you'll get 23 + 22 + 7 + 14 + 10 = 76.
If you choose the large Yeyenum, you'll get 70 + 14 + 10 + 22 = 116(all Yeyenum, since the bloggium gets blocked).

quick method count number of overlap intervals in an array of interval?

OK, this is a question I got for my advance algorithm class. I already turned in my solution once but got rejected by my instructor due to efficiency issue, in other words, I already made the efforts on my part but could not get it even after his hint, so please be gentle. I will give his hint below
Given an array of intervals with both start point and end point, find the number of other intervals fall within it for each interval. number of intervals is less than 10^9 and their ids are distinct. start and end are less than 10^18, the input files don't contain duplicate number for start and end. All the numbers above are integers
the hint is: considering a data structure with buckets. The algorithm should be faster than O(n^2)
sample input and output
input:
5 %% number of intervals
2 100 200 %% id, start,end. all lines below follows this
3 110 190
4 105 145
1 90 150
5 102 198
output:
3 0
4 0
1 1
5 2
2 3
The numbers are pretty big so O(N log N) might be a little to much but here's an idea.
First things first normalize the values, that means turning them smaller while keepinging the same ordering. In your example the normalize would be
90 100 102 105 110 145 150 190 198 200
1 2 3 4 5 6 7 8 9 10
So you're new intervals are:
5
2 2 10
3 5 8
4 4 6
1 1 7
5 3 9
Now the edges of the intervals are in the range of [1, 2N].
Now sort the intervals by their end:
5
4 4 6
1 1 7
3 5 8
5 3 9
2 2 10
When you reach an interval you can say that all the intervals that start before it and have not been encountered yet should have their answer increased by one. This can be done with a SegmentTree.
What you do when you get an interval [x, y] you increase all values in the range [1, x - 1] by 1 and then compute its answer as the value at x in the segment tree. That's just addition on an interval and query on a point, a common segment tree problem.
I don't really think you can solve this problem with less than O(N log N) time and O(N) memory, so this solution should be the asymptotically best solution in both time and space.

Hungarian algorithm - assign systematically

I'm implementing the Hungarian algorithm in a project. I managed to get it working until what is called step 4 on Wikipedia. I do manage to let the computer create enough zeroes so that the minimal amount of covering lines is the amount of rows/columns, but I'm stuck when it comes to actually assign the right agent to the right job. I see how I could assign myself, but that's more trial and error - i.e., I do not see the systematic method which is of course essential for the computer to get it work.
Say we have this matrix in the end:
a b c d
0 30 0 0 0
1 0 35 5 0
2 60 5 0 0
3 0 50 35 40
The zeroes we have to take to have each agent assigned to a job are (a, 3), (b, 0), (c,2) and (d,1). What is the system behind chosing these ones? My code now picks (b, 0) first, and ignores row 0 and column b from now on. However, it then picks (a, 1), but with this value picked there is no assignment possible for row 3 anymore.
Any hints are appreciated.
Well, I did manage to solve it in the end. The method I used was to check whether there are any columns/rows with only one zero. In such case, that agent must use that job, and that column and row have to be ignored in the future. Then, do it again so as to get a job for every agent.
In my example, (b, 0) would be the first choice. After that we have:
a b c d
0 x x x x
1 0 x 5 0
2 60 x 0 0
3 0 x 35 40
Using the method again, we can do (a, 3), etc. I'm not sure whether it has been proven that this is always correct, but it seems it is.

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