The k-th Largest Group

Time Limit: 2000MS	Memory Limit: 131072K
Total Submissions: 7869	Accepted: 2534

Description

Newman likes playing with cats. He possesses lots of cats in his home. Because the number of cats is really huge, Newman wants to group some of the cats. To do that, he first offers a number to each of the cat (1, 2, 3, …, n

). Then he occasionally combines the group cat i is in and the group cat j is in, thus creating a new group. On top of that, Newman wants to know the size of the k-th biggest group at any time. So, being a friend of Newman, can you help him?

Input

1st line: Two numbers N and M (1 ≤ N, M ≤ 200,000), namely the number of cats and the number of operations.

2nd to (m + 1)-th line: In each line, there is number C specifying the kind of operation Newman wants to do. If C = 0, then there are two numbers i and j (1 ≤ i, j ≤ n) following indicating Newman wants to combine the group containing the two cats (in case these two cats are in the same group, just do nothing); If C

= 1, then there is only one number k (1 ≤ k ≤ the current number of groups) following indicating Newman wants to know the size of the k-th largest group.

Output

For every operation “1” in the input, output one number per line, specifying the size of the kth largest group.

Sample Input

10 10
0 1 2
1 4
0 3 4
1 2
0 5 6
1 1
0 7 8
1 1
0 9 10
1 1

Sample Output

1
2
2
2
2

Hint

When there are three numbers 2 and 2 and 1, the 2nd largest number is 2 and the 3rd largest number is 1.

Source

題目大意：n只貓，編號1~n，m個操作，0 i j意思是把i在的組併到j，1 k的意思是求第k大組的大小

註釋的是我的思路，這個寫法是參照別人的思路，他的比我的稍微快個100來ms，

ac程式碼

#include<stdio.h>
#include<string.h>
int pre[200020],num[200020];
int node[200020<<2],n,m;
void build_tr(int l,int r,int tr)
{
	if(l==1)
		node[tr]=n;
	else
		node[tr]=0;
	if(l==r)
	{
		/*if(l==1)
			node[tr]=n;
		else
			node[tr]=0;*/
		return;
	}
	int mid=(l+r)>>1;
	build_tr(l,mid,tr<<1);
	build_tr(mid+1,r,tr<<1|1);
//	node[tr]=node[tr<<1]+node[tr<<1|1];
}
void init(int n)
{
	int i;
	for(i=1;i<=n;i++)
	{
		num[i]=1;
		pre[i]=i;
	}
}
int find(int x)
{
	if(x==pre[x])
		return x;
	return pre[x]=find(pre[x]);
}
void update(int pos,int add,int l,int r,int tr)
{
	node[tr]+=add;
	if(l==r)
	{
	//	node[tr]+=add;
		return;
	}
	int mid=(l+r)>>1;
	if(pos<=mid)
	{
		update(pos,add,l,mid,tr<<1);
	}
	else
		update(pos,add,mid+1,r,tr<<1|1);
	//node[tr]=node[tr<<1]+node[tr<<1|1];
}
int query(int k,int l,int r,int tr)
{
	if(l==r)
	{
		return l;
	}
	int mid=(l+r)>>1;
	if(node[tr<<1|1]>=k)
		query(k,mid+1,r,tr<<1|1);
	else
		query(k-node[tr<<1|1],l,mid,tr<<1);
	
}
int main()
{
//	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		build_tr(1,n,1);
		init(n);
		int i;
		for(i=0;i<m;i++)
		{
			int op;
			scanf("%d",&op);
			if(!op)
			{
				int a,b;
				scanf("%d%d",&a,&b);
				int fa=find(a),fb=find(b);
				if(fa!=fb)
				{
					pre[fa]=fb;
					update(num[fa],-1,1,n,1);
					update(num[fb],-1,1,n,1);
					update(num[fa]+num[fb],1,1,n,1);
					num[fb]+=num[fa];
				}
			}
			else
			{
				int a;
				scanf("%d",&a);
				printf("%d\n",query(a,1,n,1));
			}
		}
	}
}