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)
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.
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 CodeChef: Practical coding for everyone … any help is seriously welcomed
i don’t know about how the test cases were designed but still if this algorithm is implemented with union by rank it will give better performance.I am a bit disappointed that my code that should have been accepted with better performance gave TLE if anybody can explain the reason for this do reply…
Even taking only input without processing anything it gives TLE.Why this happened for java Language. It run smoothly for scanf in c if it fastio problem then it should give TLE for scanf also.
This is my
[1] in which only input is getting take by program and for this it is giving **TLE**.
[1]:http://www.codechef.com/viewsolution/4270212
I didn’t use set operations as I was worried about the time required and all that memory allocation. Simply used the transitive relations between dish owners as owner(a)–>owner(b). Simple and alternate solution.
I tried to follow the pseudo-code given above and implement the code for UF… But I’m getting WA… Can anyone tell me if something is very, very wrong here?
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100005
vector<int> parent(MAXN);
vector<int> S(MAXN);
int query,x,y,N,Q,px,py;
int find(int i)
{
int j;
if(parent[i]==i)
return i;
else
{
j=find(parent[i]);
parent[i]=j;
return j;
}
}
void 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;
}
void init(int N)
{
S.resize(N);
parent.resize(N);
for(int i = 0; i < N;i++)
parent[i]=i;
}
void solve(int query)
{
if(query == 0)
{
cin >> x >> y;
x--; y--;
px = find(x);
py = find(y);
if(px == py)
puts("Invalid query!");
else
set_union(px,py);
}
else
{
cin >> x;
x--;
cout << find(x) << endl;
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while(t--)
{
cin >> N;
init(N);
for(int i = 0; i < N; i++)
{
cin >> S[i];
}
cin >> Q;
while(Q--)
{
cin >> query;
solve(query);
}
}
return 0;
}
@kuruma thanks for this question i learnt something new.
first tiny mistake is the output of second query type you should print find(x)+1 as your numbers are 0-based .
secondly you have used “ios_base::sync_with_stdio(false)” that seems to cause your output to be buffered and is printed together only when the program terminates.
this would have been correct had you used puts/printf in output of 2nd query as well. since you used cout that output is not buffered and printed instantly while the output of all first queries is printed in the end.
So what you can do:
1.use fflush(stdout) after puts statement
2.use cout
3.NOT use ios_base::sync_with_stdio(false);
4.use puts/printf in the output of 2nd query as well
