Problem:
Disjoint Set

Write a program which manipulates a disjoint set S = {S1, S2, . . . , Sk}.

First of all, the program should read an integer n, then make a disjoint set where each element consists of 0, 1, ... n−1 respectively.

Next, the program should read an integer q and manipulate the set for q queries. There are two kinds of queries for different operations:

unite(x, y): unites sets that contain x and y, say Sx and Sy, into a new set.

same(x, y): determine whether x and y are in the same set.

Input

n q
com1 x1 y1
com2 x2 y2
...
comq xq yq

In the first line, n and q are given. Then, q queries are given where com represents the type of queries. '0' denotes unite and '1' denotes same operation.

Output

For each same operation, print 1 if x and y are in the same set, otherwise 0, in a line.

Constraints

1 ≤ n ≤ 10000

1 ≤ q ≤ 100000

x ≠ y

Sample Input

5 12
0 1 4
0 2 3
1 1 2
1 3 4
1 1 4
1 3 2
0 1 3
1 2 4
1 3 0
0 0 4
1 0 2
1 3 0

Sample Output

0
0
1
1
1
0
1
1

Solution in C++ (judge verdict: Accepted)

#include<iostream>

using namespace std;

#define MAX_N 10000

int par[MAX_N]; // ???
int rank[MAX_N]; // ??¨?????±???
// ?????????
void init(int n){
  for(int i = 0; i < n; i++){
    par[i] = i;
    rank[i] = 0;
  }
}
// ??¨??????????±???????
int find(int x){
  if(par[x] == x)
    return x;
  else
    return par[x] = find(par[x]);
}
// x??¨y????±??????????????????????
void unite(int x, int y){
  x = find(x);
  y = find(y);
  if(x == y) return;
  
  // ??¨??????????????????????????£??¨??????????????????
  if(rank[x] < rank[y]){
    par[x] = y;
  }else{
    par[y] = x;
    if(rank[x] == rank[y]) rank[x]++;
  }
}
// x??¨y?????????????????????true
bool same(int x,int y){
  return find(x) == find(y);
}

int main(){

  int n,q;
  cin >> n >> q;
  init(n);
  bool t[MAX_N];

  for(int i = 0;i < q;i++){
    int a,x,y;
    cin >> a >> x >> y;
    if(a == 0){
      unite(x,y);
    }
    else{
      bool u;
      u = same(x,y);
      if(u == true){
	cout << "1" << endl;
      }
      else{
	cout << "0" << endl;
      }
    }
  }

  

}