CHEFGAME - Editorial

PROBLEM LINK:

Practice
Contest

Author: Roman Rubanenko
Tester: Hiroto Sekido
Editorialist: Anton Lunyov

DIFFICULTY:

MEDIUM

PREREQUISITES:

Prim’s algorithm

PROBLEM:

The best described in the problem statement.

QUICK EXPLANATION:

It is clear that we have a complete weighted graph where vertices are the points and weight of the edge is the square of the distance between points. The problem asks to find the cycle-free set of edges with maximal product of weights. At first sight it seems to be the maximum spanning tree in this graph. But since some edges can have zero weights we actually should maximize the product of non-zero edges in our spanning tree. Since graph is complete then naive implementation of Prim’s algorithm would be the simplest way here.

Note that Prim’s algorithm works in O(V * V) time while Kruskal algorithm in O(E * log E) time, where V is the number of vertexes in the graph and E is the number of edges. In our case the graph is complete, so E = V * (V − 1) / 2 and hence Kruskal algorithm is much slower (due to log E factor). It was intended that Kruskal should get TLE, while any implementation of Prim (with precalculating all edges or not) will get AC.

The main bottleneck of the Kruskal algorithm is sorting of edges (not DSU stuff) and in the Prim we have no sorting at all. That is the point.

EXPLANATION:

As noted above we can reformulate the problem as follows - find the spanning tree in constructed graph that has maximum possible product of non-zero edges in it. It is well-known that spanning tree found by general greedy algorithm (like Prim’s or Kruskal’s) will maximize any symmetric monotone function of edges (I read this remark 10 years ago in Christofides book and still remember this). The product of non-zero edges is exactly one of such functions since we have no negative edges. Hence well-known algorithms for finding maximum spanning tree will work here.

The implementation of Prim’s algorithm suited for this problem is provided below. Once maximum distance between marked and not marked vertexes becomes zero we could break from the cycle. Also we do not create the adjacency matrix and calculate each needed distance on the fly.

And the most important thing is - the squares of distances should be calculated and compared as is, take them modulo 747474747 only when you multiply them to the answer. For example, if we have only two points, and square of distance between them is 747474747 then in the case of taking this distance modulo at the beginning you break from the cycle and the answer will be 1, but the correct answer is 747474747 and you should output 0. The concrete example is

2 3
0 0 0
19265 19189 2849

The output should 0.

And finally the code snippet:

input points // they are numbered from 1 to N
// d* is the farthest distance from point i to the vertexes of the tree
fill d[1..N] by zeros // should be 64bit integer type
// vertex is marked if it is in the current tree
fill mark[1..N] by false // so initially tree is empty
mod = 747474747
ans = 1 // could be int
for iter = 1 to N do
   j = 0 // will contain the farthest vertex from current tree
   for i = 1 to N do
      // we update j once we meet not marked vertex with larger distance to the tree
      if (not mark* and (j = 0 or d[j] < d*)) then
         j=i
   mark[j] = true // we add j to the tree

   // now we update the answer
   // when iter = 1 we create tree of one vertex
   // so d[j] does not make sense
   if iter > 1 then
      if d[j] <= 1 then
         break
      else
         // note that we take d[j] modulo mod
         // otherwise 64bit type overflow is possible
         ans = d[j] % mod * ans % mod

    // now we should update array d[]
    // clearly we need to update each d* for not marked i
    // by distance to j since this is the only new vertex in the tree
    for i = 1 to N do
       if not mark* then
          dist = square of distance between a* and a[j]
          // use 64bit type here but don't use modulo or the order of edges will be broken
          d* = max(d*, dist)

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution will be provided soon.
Tester’s solution can be found here.

RELATED PROBLEMS:

SPOJ - DAVIDG - 11443. Davids Greed

7 Likes

Hm, I used algorithm for spanning tree and got TLE (tried Java and C++), any idea what is inefficient in my code?

http://www.codechef.com/viewsolution/2007442

1 Like

can you plz provide me a case where my solution fails,
http://www.codechef.com/viewsolution/2052988

Me too used Kruskal’s algorithm with ranked-union, path compression, everything!! Only to find later that the sorting of edges is taking too much time :frowning: (The complete solution took ~24 seconds for a large test case, on my local machine.)

Link to Solution: http://www.codechef.com/viewplaintext/2045584

PS: Never thought Prim’s algorithm can do the trick!!

1 Like

@anton_lunyov I tried with kruskal and got TLE in edge distance calculation and sorting. I thought of prim’s algorithm too but don’t you think all the edge distances will have to be calculated in prim’s too? :o … I was getting TLE because for kruskal I calculated distance between every pair of vertices and then sorted according to the distance. Please clear my doubts. and pleasse refer to my solution and if possible tell me the bad places because of which i got TLE. Thanks in advance

Isn’t time complexity for Prims and Kruskals same : O(ElogV) where E = V^2 for complete graph.
It may be the case that prims is faster than kruskals because it doesn’t require all the edges to be sorted before hand, right?

Similar Problem:

www.spoj.pl/problems/DAVIDG

4 Likes

Can delaunay triangulation be used here to bring down the complexity to O(VlogV)???

Hi programmers,

I tried to implement Prim’s algorithm, but I’m getting WA.
I compared the results with my previous Kruskal implementation and I’m getting same result. Is there some corner case, that’s not covered by my random tests (or any other bug)?

My submission is here - http://www.codechef.com/viewsolution/2056473

Thanks

can anyone explain NZEC ?

Hello @all,

I have tried to implement Prim’s Algorithm…but I am getting TLE,I have used a stl::priority queue for the finding the min distance…could you all have a look into my program and tell me where my program is getting slow.

ideone link: http://www.codechef.com/viewsolution/2038046

Thanks in advance.

can anyone tell where this code fails i have tried many test cases it is giving correct answer but still getting wa on the judge my submission id is : http://www.codechef.com/viewsolution/2060320

I couldn’t determine why this submission of mine was failing. Please help.

http://www.codechef.com/viewsolution/2035885

Well I also can’t get it right. I always get wrong answer. Could anyone help me? :slight_smile:

My “solution”

Hey I have tried implementing the same solution as mentioned in the editorial - http://www.codechef.com/viewsolution/2170014

I am getting WA. Can anyone point out the mistake in this solution?

Thanks.

I had a doubt regarding the time complexity of Prim’s algorithm. Isn’t Prim without a heap and using adjacency list O(V*E) because the outer loop is O(V) and the inner that computes the greedy minimum values O(E).Won’t the simple implementation give TLE in this case?

Getting WA.But am not able to work out any corner cases.Can somebody give me some test cases.Help?


[1]


  [1]: http://www.codechef.com/viewsolution/2285768

I am getting WA in my approach for a similar problem DAVIDG on spoj

Can you help me figure out the mistake ?
It is on similar lines.
My Code

What happen when 2 towns are in 1 same points? should it output 0/1, as i tested in your program it gave 1, but shouldn’t it 0 because we need to connect those two town with distance of 0?

Why my solution is giving WA . I am following the approach given in editorial.
My Solution