 ### C++ Problem On Disjoint Data Structures

Problem on Disjoint Data Structures:

Owl Arena is hosting a fight competition and many owls decided to take part in it.

Strength of an owl is the number associated with that owl.

Rules of the competition are:

• An owl wins if its strength is greater than that of another.
• An owl can ask his friend to fight that match for him.

Note : If A and B are friends, and B and C are friends, then A and C are also friends.

Input:
First line contains the number of owls participating N and the number of connections M.
M lines follow.
Each line contains two integers u and v denoting that they are friends.
Next line contains Q, the number of queries.
Q lines follow.
Each line contains two integers u and v. You need to tell who wins.

Output:
In each query, output the number of the owl that will win the match. If the owls(u and v) are in the same friend circle, output $TIE$.

Constraints:
1 $%u2264$ N,M,Q $%u2264$ $10$5
u,v $%u2264$ N

Problem Setter

SAMPLE INPUT

7 31 23 41 741 25 63 71 5
SAMPLE OUTPUT

TIE671
Explanation

1,2 and 7 are friends. 3 and 4 are friends. 5,6 and 7 have no friends. now,
First query: 1 and 2 : since both belong to the same friend circle, answer is a TIE.
Second: 5 and 6: since there is no friend of 5 and no friend of 6 and 6 has higher strength. 6 will win.
Third: 3 and 7: 3 has a friend 4 who has more strength than 3 and 7 has no friends whose strength is greater than his. so 4 vs 7. 7 will win.
Fourth: 1 and 5: 1 has a friend 7 who has more strength than 1 and 5 has no friends. so 5 vs 7. 7 will win. And since the fight was b/w 1 and 5. 1 wins the fight.

Implementation code for above problem:

#include<iostream>
using namespace std;
//For parent array
int a;
int find(int f){
//if it is a parent
if(a[f]<0){
return f;
}
else{
return a[f]=find(a[f]);
}
}
void Union(int a1,int b1){
//assign element with larger value as parent to another
a[a1]=min(a[a1],a[b1]);
a[b1]=a1;
}
int main(){
int n,k;
cin>>n>>k;
for(int i=1;i<=n;i++){
a[i]=-i;
}
int a1,b1;
for(int i=1;i<=k;i++){
cin>>a1>>b1;
int para=find(a1);
int parb=find(b1);
if(para!=parb){
Union(para,parb);
}
}
cin>>k;
for(int i=1;i<=k;i++){
cin>>a1>>b1;
int para=find(a1);
int parb=find(b1);
if(para==parb){
cout<<"TIE"<<endl;
}
else if(a[para]<a[parb]){
cout<<a1<<endl;
}
else{
cout<<b1<<endl;
}
}
}
---------------------------------------------------
Input:

7 3
1 2
3 4
1 7
4
1 2
5 6
3 7
1 5

Output:

TIE
6
7
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