There was a recent contest on codechef named CODEMASTER (which has just ended a few minutes back, so I can put this question on a forum now, I believe :P ).
The question is Archers and Pikemen.
This is the problem statement::
You are being attacked by hostile enemy forces, with knights and
swordsmen charging at you. Being the commander of your unit, you have
been given the task of arranging your elite archers and pikemen in a
special formation.
The archers and pikemen must stand between two flag posts in a
straight line (no soldier can stand beyond the flags). Each archer
must have at least two pikemen by his side (one to his left and one to
his right) such that he is at equal distances from both of them. (A
pikeman may be shared between two archers).
The archers stand at given fixed positions and the separations between
them may not be equal. You need to position your troops in the given
formation using the minimum number of pikemen.
Assume that the minimum distance between a pikeman and an archer or a
pikeman and a flag is 1 unit. the minimum distance between two
archers is 2 units. Input
The first line of the input contains an integer T denoting the number
of test cases. The description of T test cases follow.
The second line contains an integer N denoting the number of
separations.
The following N lines each contain an integer x, which is the
separation of the current archer from the previous one.
The first value of x is the separation of the first archer from the
first flag. The last value of x is the separation between the last
archer and the second flag Output
For each test case, output a single line containing the minimum number
of pikemen required. Constraints
1 ≤ T ≤ 1000
1 ≤ N ≤ 1000
2 ≤ x ≤ 1000
Example
Input:
2 3 4 4 2 4 2 2 2 2
Output:
3 4
Explanation
Example case 1: A possible formation can be :
F---1---p---3---A---3---p---1---A---1---p---1--- F
Example case 2: A possible formation can be :
F---1---p---1---A---1---p---1---A---1---p---1---A---1---p---1---F
F = flag A = archer p = pikeman
---d--- = distance between pikeman and archer/flag
The first instinct was that the number of pikeman were equal to the number of gaps, but then I realized that there can be a case when, we might that to place two pikeman between 2 archers since the distance of the next archer to the right might have their distance less than the distance between previous two archers.
Can some please help me explain the algorithm for this question.
The problem link:: http://www.codechef.com/CDMS2014/problems/CM1401
Link to one of the accepted solutions:: http://www.codechef.com/viewsolution/5166485
Please help me explain this problem guys.
Thanks in advance.. ;)
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'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
I'm making some exercises on combinatorics algorithm and trying to figure out how to solve the question below:
Given a group of 25 bits, set (choose) 15 (non-permutable and order NON matters):
n!/(k!(n-k)!) = 3.268.760
Now for every of these possibilities construct a matrix where I cross every unique 25bit member against all other 25bit member where
in the relation in between it there must be at least 11 common setted bits (only ones, not zeroes).
Let me try to illustrate representing it as binary data, so the first member would be:
0000000000111111111111111 (10 zeros and 15 ones) or (15 bits set on 25 bits)
0000000001011111111111111 second member
0000000001101111111111111 third member
0000000001110111111111111 and so on....
...
1111111111111110000000000 up to here. The 3.268.760 member.
Now crossing these values over a matrix for the 1 x 1 I must have 15 bits common. Since the result is >= 11 it is a "useful" result.
For the 1 x 2 we have 14 bits common so also a valid result.
Doing that for all members, finally, crossing 1 x 3.268.760 should result in 5 bits common so since it's < 11 its not "useful".
What I need is to find out (by math or algorithm) wich is the minimum number of members needed to cover all possibilities having 11 bits common.
In other words a group of N members that if tested against all others may have at least 11 bits common over the whole 3.268.760 x 3.268.760 universe.
Using a brute force algorithm I found out that with 81 25bit member is possible achive this. But i'm guessing that this number should be smaller (something near 12).
I was trying to use a brute force algorithm to make all possible variations of 12 members over the 3.268.760 but the number of possibilities
it's so huge that it would take more than a hundred years to compute (3,156x10e69 combinations).
I've googled about combinatorics but there are so many fields that i don't know in wich these problem should fit.
So any directions on wich field of combinatorics, or any algorithm for these issue is greatly appreciate.
PS: Just for reference. The "likeness" of two members is calculated using:
(Not(a xor b)) and a
After that there's a small recursive loop to count the bits given the number of common bits.
EDIT: As promissed (#btilly)on the comment below here's the 'fractal' image of the relations or link to image
The color scale ranges from red (15bits match) to green (11bits match) to black for values smaller than 10bits.
This image is just sample of the 4096 first groups.
tl;dr: you want to solve dominating set on a large, extremely symmetric graph. btilly is right that you should not expect an exact answer. If this were my problem, I would try local search starting with the greedy solution. Pick one set and try to get rid of it by changing the others. This requires data structures to keep track of which sets are covered exactly once.
EDIT: Okay, here's a better idea for a lower bound. For every k from 1 to the value of the optimal solution, there's a lower bound of [25 choose 15] * k / [maximum joint coverage of k sets]. Your bound of 12 (actually 10 by my reckoning, since you forgot some neighbors) corresponds to k = 1. Proof sketch: fix an arbitrary solution with m sets and consider the most coverage that can be obtained by k of the m. Build a fractional solution where all symmetries of the chosen k are averaged together and scaled so that each element is covered once. The cost of this solution is [25 choose 15] * k / [maximum joint coverage of those k sets], which is at least as large as the lower bound we're shooting for. It's still at least as small, however, as the original m-set solution, as the marginal returns of each set are decreasing.
Computing maximum coverage is in general hard, but there's a factor (e/(e-1))-approximation (≈ 1.58) algorithm: greedy, which it sounds as though you could implement quickly (note: you need to choose the set that covers the most uncovered other sets each time). By multiplying the greedy solution by e/(e-1), we obtain an upper bound on the maximum coverage of k elements, which suffices to power the lower bound described in the previous paragraph.
Warning: if this upper bound is larger than [25 choose 15], then k is too large!
This type of problem is extremely hard, you should not expect to be able to find the exact answer.
A greedy solution should produce a "fairly good" answer. But..how to be greedy?
The idea is to always choose the next element to be the one that is going to match as many possibilities as you can that are currently unmatched. Unfortunately with over 3 million possible members, that you have to try match against millions of unmatched members (note, your best next guess might already match another member in your candidate set..), even choosing that next element is probably not feasible.
So we'll have to be greedy about choosing the next element. We will choose each bit to maximize the sum of the probabilities of eventually matching all of the currently unmatched elements.
For that we will need a 2-dimensional lookup table P such that P(n, m) is the probability that two random members will turn out to have at least 11 bits in common, if m of the first n bits that are 1 in the first member are also 1 in the second. This table of 225 probabilities should be precomputed.
This table can easily be computed using the following rules:
P(15, m) is 0 if m < 11, 1 otherwise.
For n < 15:
P(n, m) = P(n+1, m+1) * (15-m) / (25-n) + P(n+1, m) * (10-n+m) / (25-n)
Now let's start with a few members that are "very far" from each other. My suggestion would be:
First 15 bits 1, rest 0.
First 10 bits 0, rest 1.
First 8 bits 1, last 7 1, rest 0.
Bits 1-4, 9-12, 16-23 are 1, rest 0.
Now starting with your universe of (25 choose 15) members, eliminate all of those that match one of the elements in your initial collection.
Next we go into the heart of the algorithm.
While there are unmatched members:
Find the bit that appears in the most unmatched members (break ties randomly)
Make that the first set bit of our candidate member for the group.
While the candidate member has less than 15 set bits:
Let p_best = 0, bit_best = 0;
For each unset bit:
Let p = 0
For each unmatched member:
p += P(n, m) where m = number of bits in common between
candidate member+this bit and the unmatched member
and n = bits in candidate member + 1
If p_best < p:
p_best = p
bit_best = this unset bit
Set bit_best as the next bit in our candidate member.
Add the candidate member to our collection
Remove all unmatched members that match this from unmatched members
The list of candidate members is our answer
I have not written code, I therefore have no idea how good an answer this algorithm will produce. But assuming that it does no better than your current, for 77 candidate members (we cheated and started with 4) you have to make 271 passes through your unmatched candidates (25 to find the first bit, 24 to find the second, etc down to 11 to find the 15th, and one more to remove the matched members). That's 20867 passes. If you have an average of 1 million unmatched members, that's on the order of a 20 billion operations.
This won't be quick. But it should be computationally feasible.
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.