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 JAVA (judge verdict: Accepted)


import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = Integer.parseInt(scan.next());
        int q = Integer.parseInt(scan.next());

        UnionFind uf = new UnionFind(n);
        for (int i = 0; i < q; i++) {
            int com = Integer.parseInt(scan.next());
            int x = Integer.parseInt(scan.next());
            int y = Integer.parseInt(scan.next());
            if (com == 0) {
                uf.unite(x, y);
            } else {
                if (uf.same(x, y)) {
                    System.out.println("1");
                } else {
                    System.out.println("0");
                }
            }
        }
        scan.close();
    }

    static class UnionFind {
        /** ルートノード */
        private int[] roots;

        /**
         * コンストラクタ.
         *
         * @param n ノード数
         */
        public UnionFind(int n) {
            roots = new int[n];
            for (int i = 0; i < n; i++) {
                roots[i] = i;
            }
        }

        /**
         * 各々のノードが属するグループを結合する.
         *
         * @param x
         * @param y
         */
        public void unite(int x, int y) {
            roots[getRoot(y)] = getRoot(x);
        }

        /**
         * 各々のノードが属するグループが同じかどうかを判定する.
         *
         * @param x
         * @param y
         * @return 同じグループに属するならtrue
         */
        public boolean same(int x, int y) {
            return getRoot(x) == getRoot(y);
        }

        /**
         * ルートノードを返す.<br>
         * 再帰によって経路圧縮を行っている.
         *
         * @param x
         * @return ルートノード
         */
        private int getRoot(int x) {
            int parent = roots[x];
            if (x != parent) {
                roots[x] = getRoot(parent);
            }
            return roots[x];
        }
    }

}