HDU1856 (More is better)-query set + tree

[Java]
Package D0813;
/*
* Questions: mr wang puts a group of children in a room. Some of these children are friends,
* Mr wang selects some children from the room each time and asks you at most a few children (the children must be friends,
* It can be either direct or inter-node), giving a direct friend relationship, and asking you to determine the number of children left most.
* Train of Thought: Check the set and find the result with the largest number of sets.
**/
Import java. io .*;
Public class HDU1856 {
Static int [] set;
Static int [] height;
Static int [] boy; // number of nodes.
Static int max;
Public static void main (String [] args) throws IOException {
StreamTokenizer st = new StreamTokenizer (new BufferedReader (new InputStreamReader (System. in )));
While (st. nextToken ()! = StreamTokenizer. TT_EOF ){
Int n = (int) st. nval;
Init (n );
Max = 1; // be sure to pay attention to this place. Do not set it to 0. Because at least one child can be left in the end.
For (int I = 0; I <n; I ++ ){
St. nextToken ();
Int s = (int) st. nval;
St. nextToken ();
Int e = (int) st. nval;
Merge (s, e );
}
System. out. println (max );
}
}
 
Public static void init (int n ){
Boy = new int [2 * n + 1];
Height = new int [2 * n + 1];
Set = new int [2 * n + 1];
For (int I = 1; I <= 2 * n; I ++ ){
Set [I] = I;
Height [I] = 1;
Boy [I] = 1;
}
}
Public static int find (int x ){
Int son, temp;
Son = x;
While (x! = Set [x])
X = set [x];
// Path compression, which can be left empty
While (son! = X ){
Temp = set [son];
Set [son] = x;
Son = temp;
}
Return x;
}
Public static void merge (int a, int B ){
A = find ();
B = find (B );
If (! = B ){
If (height [a]> height [B]) {
Set [B] =;
Boy [a] + = boy [B];
Max = Math. max (max, boy [a]);
}
Else if (height [a] Set [a] = B;
Boy [B] + = boy [a];
Max = Math. max (max, boy [B]);
}
Else {
Set [B] =;
Height [a] ++;
Boy [a] + = boy [B];
Max = Math. max (max, boy [a]);
}
}
}
}

 

Author; lhfight