Disjoint-set template title ---- P3367 [template] disjoint-set

Title Description

As stated, there is now a disjoint-set, you need to complete the merge and query operations.

Input Format

The first line contains two integers N, M, N represents a total of M elements and operations.

Next M rows, each row containing three integers Zi, Xi, Yi

When Zi = 1, the combined set of Xi and Yi where

When Zi = 2, the output Xi and Yi are in the same set, is if the Y output; otherwise it outputs N

Output Format

As above, the operation for each of the Zi = 2, has output line, each row contains a capital letter, or Y is N

Sample input and output

Input # 1copy

4 7
2 1 2
1 1 2
2 1 2
1 3 4
2 1 4
1 2 3
2 1 4

Output # 1copy

N 
Y 
N 
Y

Description / Tips

Constraints of time: 1000ms, 128M

Data Scale:

For 30% of the data, N <= 10, M <= 20;

For 70% of the data, N <= 100, M <= 1000;

To 100% of the data, N <= 10000, M <= 200000.

 

Special attention: fa array needs to be initialized, the ancestors of every node are their own

AC codes and templates

#include <the iostream>
 the using  namespace STD;
 const  int NN = 1e6 + 10 ;
 int N, M;
 int FA [NN]; 

int Find ( int X) { // query and compression path 
    IF (X =! FA [X] ) 
        FA [X] = Find (FA [X]);
     return FA [X]; 
} 

void the Join ( int x, int y) { // combined x, y set where 
    int FX = Find (X), FY = Find (Y);
     IF (! FX = FY) 
        FA [FX] =FY; 
} 

int main () { 
    CIN >> N >> M;
     for ( int I = . 1 ; I <= N; I ++) FA [I] = I; // initialize each node is its own ancestor 
    int OP , T1, T2;
     the while (M-- ) { 
        Scanf ( " % D% D% D " , & OP, & T1, & T2);
         IF (OP == . 1 ) { 
            the Join (T1, T2); // merge 
        } the else {
             iF (Find (T1) == Find (T2)) { // query is the same ancestor 
                COUT << " the Y"<<endl;
            }else{
                cout<<"N"<<endl;
            }
        }
    }

}