I'm currently trying to solve an algorithm problem from last year's Polish Collegiate Championships which reads as follows:
The Lord Mayor of Bytetown plans to locate a number of radar speed
cameras in the city. There are n intersections in Bytetown numbered
from 1 to n, and n-1 two way street segments. Each of these street
segments stretches between two intersections. The street network
allows getting from each intersection to any other.
The speed cameras are to be located at the intersections (maximum one
per intersection), wherein The Lord Mayor wants to maximise the number
of speed cameras. However, in order not to aggravate Byteland
motorists too much, he decided that on every route running across
Bytetown roads that does not pass through any intersection twice there
can be maximum k speed cameras (including those on endpoints of the
route). Your task is to write a program which will determine where the
speed cameras should be located.
Input
The first line of input contains two integers n and k (1 <= n, k <=
1000000): the number of intersections in Bytetown and maximum number
of speed cameras which can be set up on an individual route. The lines
that follow describe Bytetown street network: the i-th line contains
two integers a_i and b_i (1 <= a_i, b_i <= n), meaning that there is a
two-way street segment which joins two intersections numbered a_i and
b_i.
Output
The first output line should produce m: the number describing the
maximum number of speed cameras, that can be set up in Byteland. The
second line should produce a sequence of m numbers describing the
intersections where the speed cameras should be constructed. Should
there be many solutions, your program may output any one of them.
Example
For the following input data:
5 2
1 3
2 3
3 4
4 5
one of the correct results is:
3 1 2 4
So judging by how many teams solved it, I'm guessing it can't be too hard but still, I got stuck almost immediately with no idea as to how to move on. Since we know that "on every route running across Bytetown roads that does not pass through any intersection twice there can be maximum k speed camera", I guess we first have to somehow dissect the graphs into components being possible routes around the town. This alone seems like a really hard thing to do cause supposing there's an intersection with four motorways coming out of it, it already creates three possible directions for every enter point, thus making 12 routes. Not to mention how the situation complicates when there's more such four-handed intersections.
Maybe I'm approaching the task from the wrong angle? Could you please help?
It seems greedy works here
while k >= 2
mark all leaves of the tree and remove them
k = k - 2;
if ( k == 1 )
mark any 1 of remaining vertices
Related
Two different lists having radii of upper hemisphere and lower hemisphere is provided. The first list consists of N upper hemispheres indexed 1 to N and the second has M lower hemispheres indexed 1 to M. A sphere of radius of R can be made taking one upper half of the radius R and one lower half of the radius R. Also, you can put a sphere into a bigger one and create a sequence of nested concentric spheres. But you can't put two or more spheres directly into another one.
If there is a sequence of (D+1) nested spheres, we can call this sequence as a D-sequence.
Find out how many different X-sequence are possible (1 <= X <= C). An X sequence is different from another if the index of any of the hemisphere used in one X-sequence is different from the other.
INPUT
The first line contains a three integers: N denoting the number of upper sphere halves, M denoting the number of lower sphere halves and C.
The second line contains N space-separated integers denoting the radii of upper hemispheres.
The third line contains M space-separated integers denoting the radii of lower hemispheres.
OUTPUT
Output a single line containing C space-separated integers , the number of ways there are to build i-sequence in modulo 1000000007.
Example
Input
3 4 3
1 2 3
1 1 3 2
Output
5 2 0
I am looking for those elements which are part of both the lists of upper as well as lower hemispheres, so that they can form a sphere and then taking their maximum count by comparing their counts in both radii lists.
And, So, for different C sum of products of counts of C+1 elements yields the answer.
How to calculate the above efficiently or is there any other approach ??
Guys this is my first answer. Spare me the whip for now as i am here to learn.
You first find the numbers of spheres possible for each radii.
no of spheres: 2 1 1
Having Radii: 1 2 3
Now since we can fit a sphere with radius r inside a sphere with radii R such that R>r, all we need to do is to find the no . of increasing subsequences of length 2,3,...till c in the list of all possible spheres formed.
List of possible spheres:[1,1*,2,3](* used for marking)
consider D1: it has 2 spheres. Try finding the no. of increasing subsequences of length 2 in the above list.
They are:
[1,2],[1*,2][1,3][1*,3][2,3]
hence the ans is 5.
Get it??
Now how to solve:
It can be done by using Dp. Naive solution has complexity .O(n^2*constant).
You may follow along the lines as provided in the following link :Dp solution.
It is worth mentioning that faster methods do exist which use BIT , segment trees etc.
It is similar to this SPOJ problem.
Can anyone please suggest me algorithm for this.
You are given starting and the ending points of N segments over the x-axis.
How many of these segments can be touched, even on their edges, by exactly two lines perpendicular to them?
Sample Input :
3
5
2 3
1 3
1 5
3 4
4 5
5
1 2
1 3
2 3
1 4
1 5
3
1 2
3 4
5 6
Sample Output :
Case 1: 5
Case 2: 5
Case 3: 2
Explanation :
Case 1: We will draw two lines (parallel to Y-axis) crossing X-axis at point 2 and 4. These two lines will touch all the five segments.
Case 2: We can touch all the points even with one line crossing X-axis at 2.
Case 3: It is not possible to touch more than two points in this case.
Constraints:
1 ≤ N ≤ 10^5
0 ≤ a < b ≤ 10^9
Let assume that we have a data structure that supports the following operations efficiently:
Add a segment.
Delete a segment.
Return the maximum number of segments that cover one point(that is, the "best" point).
If have such a structure, we can get use the initial problem efficiently in the following manner:
Let's create an array of events(one event for the start of each segment and one for the end) and sort by the x-coordinate.
Add all segments to the magical data structure.
Iterate over all events and do the following: when a segment start, add one to the number of currently covered segments and remove it from that data structure. When a segment ends, subtract one from the number of currently covered segment and add this segment to the magical data structure. After each event, update the answer with the value of the number of currently covered segments(it shows how many segments are covered by the point which corresponds to the current event) plus the maximum returned by the data structure described above(it shows how we can choose another point in the best possible way).
If this data structure can perform all given operations in O(log n), then we have an O(n log n) solution(we sort the events and make one pass over the sorted array making a constant number of queries to this data structure for each event).
So how can we implement this data structure? Well, a segment tree works fine here. Adding a segment is adding one to a specific range. Removing a segment is subtracting one from all elements in a specific range. Get ting the maximum is just a standard maximum operation on a segment tree. So we need a segment tree that supports two operations: add a constant to a range and get maximum for the entire tree. It can be done in O(log n) time per query.
One more note: a standard segment tree requires coordinates to be small. We may assume that they never exceed 2 * n(if it is not the case, we can compress them).
An O(N*max(logN, M)) solution, where M is the medium segment size, implemented in Common Lisp: touching-segments.lisp.
The idea is to first calculate from left to right at every interesting point the number of segments that would be touched by a line there (open-left-to-right on the lisp code). Cost: O(NlogN)
Then, from right to left it calculates, again at every interesting point P, the best location for a line considering segments fully to the right of P (open-right-to-left on the lisp code). Cost O(N*max(logN, M))
Then it is just a matter of looking for the point where the sum of both values tops. Cost O(N).
The code is barely tested and may contain bugs. Also, I have not bothered to handle edge cases as when the number of segments is zero.
The problem can be solved in O(Nlog(N)) time per test case.
Observe that there is an optimal placement of two vertical lines each of which go through some segment endpoints
Compress segments' coordinates. More info at What is coordinate compression?
Build a sorted set of segment endpoints X
Sort segments [a_i,b_i] by a_i
Let Q be a priority queue which stores right endpoints of segments processed so far
Let T be a max interval tree built over x-coordinates. Some useful reading atWhat are some sources (books, etc.) from where I can learn about Interval, Segment, Range trees?
For each segment make [a_i,b_i]-range increment-by-1 query to T. It allows to find maximum number of segments covering some x in [a,b]
Iterate over elements x of X. For each x process segments (not already processed) with x >= a_i. The processing includes pushing b_i to Q and making [a_i,b_i]-range increment-by-(-1) query to T. After removing from Q all elements < x, A= Q.size is equal to number of segments covering x. B = T.rmq(x + 1, M) returns maximum number of segments that do not cover x and cover some fixed y > x. A + B is a candidate for an answer.
Source:
http://www.quora.com/What-are-the-intended-solutions-for-the-Touching-segments-and-the-Smallest-String-and-Regex-problems-from-the-Cisco-Software-Challenge-held-on-Hackerrank
I have encoutered an algorithm question:
Fully Connection
Given n cities which spreads along a line, let Xi be the position of city i and Pi be its population.
Now we begin to lay cables between every two of the cities based on their distance and population. Given two cities i and j, the cost to lay cable between them is |Xi-Xj|*max(Pi,Pj). How much does it cost to lay all the cables?
For example, given:
i Xi Pi
- -- --
1 1 4
2 2 5
3 3 6
Then the total cost can be calculated as:
i j |Xi-Xj| max(Pi, Pj) Segment Cost
- - ------ ----------- ------------
1 2 1 5 5
2 3 1 6 6
1 3 2 6 12
So that the total cost is 5+6+12 = 23.
While this can clearly be done in O(n2) time, can it be done in asymptotically less time?
I can think of faster solution. If I am not wrong it goes to O(n*logn). Now let's first sort all the cities according to Pi. This is O(n* log n). Then we start processing the cities in increasing order of Pi. the reason being - you always know you have max (Pi, Pj) = Pi in this case. We only want to add all the segments that come from relations with Pi. Those that will connect with larger indexes will be counted when they will be processed.
Now the thing I was able to think of was to use several index trees in order to reduce the complexity of the algorithm. First index tree is counting the number of nodes and can process queries of the kind: how many nodes are to the right of xi in logarithmic time. Lets call this number NR. The second index tree can process queries of the kind: what is the sum of distances from all the points to the right of a given x. The distances are counted towards a fixed point that is guaranteed to be to the right of the rightmost point, lets call its x XR.Lets call this number SUMD. Then the sum of the distances to all points to the right of our point can be found that way: NR * dist(Xi, XR) - SUMD. Then all these contribute (NR * dist(Xi, XR) - SUMD) *Pi to the result. The same for the left points and you get the answer. After you process the ith point you add it to the index trees and can go on.
Edit: Here is one article about Biary index trees: http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees
This is the direct connections problem from codesprint 2.
They will be posting worked solutions to all problems within a week on their website.
(They have said "Now that the contest is over, we're totally cool with everyone discussing their solutions to the problems.")
Okay here is a puzzle I come across a lot of times-
Given a set of 12 balls , one of which is defective (it weighs either less or more) . You are allow to weigh 3 times to find the defective and also tell which weighs less or more.
The solution to this problem exists, but I want to know whether we can algorithmically determine if given a set of 'n' balls what is the minimum number of times you would need to use a beam balance to determine which one is defective and how(lighter or heavier).
A wonderful algorithm by Jack Wert can be found here
http://www.cut-the-knot.org/blue/OddCoinProblems.shtml
(as described for the case n is of the form (3^k-3)/2, but it is generalizable to other n, see the writeup below)
A shorter version and probably more readable version of that is here
http://www.cut-the-knot.org/blue/OddCoinProblemsShort.shtml
For n of the form (3^k-3)/2, the above solution applies perfectly and the minimum number of weighings required is k.
In other cases...
Adapting Jack Wert's algorithm for all n.
In order to modify the above algorithm for all n, you can try the following (I haven't tried proving the correctness, though):
First check if n is of the from (3^k-3)/2. If it is, apply above algorithm.
If not,
If n = 3t (i.e. n is a multiple of 3), you find the least m > n such that m is of the form (3^k-3)/2. The number of weighings required will be k. Now form the groups 1, 3, 3^2, ..., 3^(k-2), Z, where 3^(k-2) < Z < 3^(k-1) and repeat the algorithm from Jack's solution.
Note: We would also need to generalize the method A (the case when we know if the coin is heavier of lighter), for arbitrary Z.
If n = 3t+1, try to solve for 3t (keeping one ball aside). If you don't find the odd ball among 3t, the one you kept aside is defective.
If n = 3t+2, form the groups for 3t+3, but have one group not have the one ball group. If you come to the stage when you have to rotate the one ball group, you know the defective ball is one of two balls and you can then weigh one of those two balls against one of the known good balls (from among the other 3t).
Trichotomy ! :)
Explanation :
Given a set of n balls, subdivide it in 3 sets A, B and C of n/3 balls.
Compare A and B. If equal, then the defective ball is in C.
etc.
So, your minimum number of times is the number of times you can divide n by three (sorry, i do not know the english word for that).
You could use a general planning algorithm: http://www.inf.ed.ac.uk/teaching/courses/plan/
I'm training code problems, and on this one I am having problems to solve it, can you give me some tips how to solve it please.
The problem is taken from here:
https://www.ieee.org/documents/IEEEXtreme2008_Competitition_book_2.pdf
Problem 12: Cynical Times.
The problem is something like this (but do refer to above link of the source problem, it has a diagram!):
Your task is to find the sequence of points on the map that the bomber is expected to travel such that it hits all vital links. A link from A to B is vital when its absence isolates completely A from B. In other words, the only way to go from A to B (or vice versa) is via that link.
Due to enemy counter-attack, the plane may have to retreat at any moment, so the plane should follow, at each moment, to the closest vital link possible, even if in the end the total distance grows larger.
Given all coordinates (the initial position of the plane and the nodes in the map) and the range R, you have to determine the sequence of positions in which the plane has to drop bombs.
This sequence should start (takeoff) and finish (landing) at the initial position. Except for the start and finish, all the other positions have to fall exactly in a segment of the map (i.e. it should correspond to a point in a non-hit vital link segment).
The coordinate system used will be UTM (Universal Transverse Mercator) northing and easting, which basically corresponds to a Euclidian perspective of the world (X=Easting; Y=Northing).
Input
Each input file will start with three floating point numbers indicating the X0 and Y0 coordinates of the airport and the range R. The second line contains an integer, N, indicating the number of nodes in the road network graph. Then, the next N (<10000) lines will each contain a pair of floating point numbers indicating the Xi and Yi coordinates (1 < i<=N). Notice that the index i becomes the identifier of each node. Finally, the last block starts with an integer M, indicating the number of links. Then the next M (<10000) lines will each have two integers, Ak and Bk (1 < Ak,Bk <=N; 0 < k < M) that correspond to the identifiers of the points that are linked together.
No two links will ever cross with each other.
Output
The program will print the sequence of coordinates (pairs of floating point numbers with exactly one decimal place), each one at a line, in the order that the plane should visit (starting and ending in the airport).
Sample input 1
102.3 553.9 0.2
14
342.2 832.5
596.2 638.5
479.7 991.3
720.4 874.8
744.3 1284.1
1294.6 924.2
1467.5 659.6
1802.6 659.6
1686.2 860.7
1548.6 1111.2
1834.4 1054.8
564.4 1442.8
850.1 1460.5
1294.6 1485.1
17
1 2
1 3
2 4
3 4
4 5
4 6
6 7
7 8
8 9
8 10
9 10
10 11
6 11
5 12
5 13
12 13
13 14
Sample output 1
102.3 553.9
720.4 874.8
850.1 1460.5
102.3 553.9
Pre-process the input first, so you identify the choke points. Algorithms like Floyd-Warshall would help you.
Model the problem as a Heuristic Search problem, you can compute a MST which covers all choke-points and take the sum of the costs of the edges as a heuristic.
As the commenters said, try to make concrete questions, either here or to the TA supervising your class.
Don't forget to mention where you got these hints.
The problem can be broken down into two parts.
1) Find the vital links.
These are nothing but the Bridges in the graph described. See the wiki page (linked to in the previous sentence), it mentions an algorithm by Tarjan to find the bridges.
2) Once you have the vital links, you need to find the smallest number of points which given the radius of the bomb, will cover the links. For this, for each link, you create a region around it, where dropping the bomb will destroy it. Now you form a graph of these regions (two regions are adjacent if they intersect). You probably need to find a minimum clique partition in this graph.
Haven't thought it through (especially part 2), but hope it helps.
And good luck in the contest!
I think Moron' is right about the first part, but on the second part...
The problem description does not tell anything about "smallest number of points". It tells that the plane flies to the closest vital link.
So, I think the part 2 will be much simpler:
Find the closest non-hit segment to the current location.
Travel to the closest point on the closest segment.
Bomb the current location (remove all segments intersecting a circle)
Repeat until there are no non-hit vital links left.
This straight-forward algorithm has a complexity of O(N*N), but this should be sufficient considering input constraints.