N chef’s are there. Each chef has a dish which is given a value (denoted by S[i]). You need to answer 2 queries :
0 x y : Chef containing dish x competes with chef containing dish y. If both dishes belong to same chef print “Invalid Query!” , otherwise the winner of the round will obtain all the dishes of the loser which will then be eliminated.
Union Find or Disjoint Set Data structures allow you to merge two sets of items together dynamically and maintain for each item - to which set set does it belong.
This is exactly what we need in the problem , i.e. dynamically merging two sets and querying which set does an element belong.
We will be using Disjoint Set Data Structure that supports the following :
find(A) : Get the id of the disjoint set to which A belongs(i.e. the chef id which contains that dish).
union(A,B) : Merge the two disjoint sets in which A and B belong respectively.
Initially, we will create N disjoint set , each set contains 1 dish and it’s owner.
Let dish x be in setA and dish y be in setB.
If the dish contained by the owner of setA has score higher than the dish contained by the owner of setB , then we will merge setA and setB and set the owner of setA as the overall owner of the merged set.
If the dish contained by the owner of setB has score higher than the dish contained by the owner of setA , then we will merge setA and setB and set the owner of setB as the overall owner of the merged set.
Note : You can easily prove from this that the owner of the set has dish whose score is higher than all other dish of the set.
We will be using only Path Compression heuristics to solve the problem.
Pseudo Code
Initialize parent[i] = i
Let S[i] denote the initial array.
int find(int i)
int j
if(parent[i]==i)
return i
else
j=find(parent[i])
//Path Compression Heuristics
parent[i]=j
return j
set_union(int i,int j)
int x1,y1
x1=find(i)
y1=find(j)
//parent of both of them will be the one with the highest score
if(S[x1]>S[y1])
parent[y1]=x1
else if ( S[x1] < S[y1])
parent[x1]=y1
solve()
if(query == 0)
Input x and y
px = find(x)
py = find(y)
if(px == py)
print "Invalid query!"
else
set_union(px,py)
else
Input x.
print find(x)
I have solved with the same approach but was getting TLE as i was using Console.WriteLine to print all the output in C# . When i used the StringBuilder to store the messages and output in the last it passed the test cases. I am not sure if we have to consider this small optimization also
@dev8546: Thanks for your comment. Now I understand the reason that I never got it AC. I couldn’t think that Console.WriteLine() might be causing the problem. I kept on hitting my head thinking there might be a better optimized way to solve the problem and finally gave up on it after trying some micro optimizations
Hai, I don’t know whether I can post this or not ,Here are some unofficial test cases categorised into small,medium,large.Who are getting mainly WA but also RTE,TLE,… can also check.You may need not worry about the constraints.the values are guarenteed to be in the given constraints.
Did the same in Java as dev8546 described. By printing System.out.println() it gets TLE. If i use a stringbuilder and output it at the end it passes. (I used both path compression and union by rank).
don’t use System.out.println(), instead use StringBuilder to store answers and then in the end use System.out.println() to print StringBuilder object. I have modified your code it has got accepted.
Link to ur code: http://www.codechef.com/status/DISHOWN,sousnake
(see the accepted solution).
I used UF data structure with path compression but still got TLE. Why would the program run into TLE when using System.out.println()? Can anybody explain that?
anyone would tell wats wrng in my code … its same as that of the editorial !!!
the link of the solution is http://www.codechef.com/viewsolution/4326802 … any help is seriously welcomed
