I hope you know the beautiful Union-Find structure. In this problem, you’re to implement something similar, but not identical. The data structure you need to write is also a collection of disjoint sets, supporting 3 operations:

• 1 p q Union the sets containing p and q. If p and q are already in the same set, ignore this command.
• 2 p q Move p to the set containing q. If p and q are already in the same set, ignore this command.
• 3 p Return the number of elements and the sum of elements in the set containing p.

Initially, the collection contains n sets: {1}, {2}, {3}, . . . , {n}.

There are several test cases. Each test case begins with a line containing two integers n and m (1 ≤ n, m ≤ 100, 000), the number of integers, and the number of commands. Each of the next m lines contains a command. For every operation, 1 ≤ p, q ≤ n. The input is terminated by end-of-file (EOF).

For each type-3 command, output 2 integers: the number of elements and the sum of elements.

Explanation

Initially: {1}, {2}, {3}, {4}, {5}

Collection after operation 1 1 2: {1,2}, {3}, {4}, {5}

Collection after operation 2 3 4: {1,2}, {3,4}, {5} (we omit the empty set that is produced when taking out 3 from {3})

Collection after operation 1 3 5: {1,2}, {3,4,5}

Collection after operation 2 4 1: {1,2,4}, {3,5}

5 7 
1 1 2 
2 3 4 
1 3 5 
3 4 
2 4 1 
3 4 
3 3

3 12 
3 7 
2 8

题意：

给出1-N的数，一开始每个数自形成一个集合，要求支持一下操作

• 将元素 p 所在集合与元素 q 所在集合合并
• 将元素 p 移到元素 q 所在集合
• 询问元素 p 所在集合有多少个元素，和为多少

题解:

第一和第三个操作都是很裸的并查集，关键在于操作2，很容易可以想到，元素 p 移到元素 q所在集合，可以先在元素 p 所在集合将其删除，然后把元素 p 与元素 q 所在集合合并。而这一步的关键在于删除操作，如果 p 是叶子节点，直接改变 p 的父亲指向就行，但是 p 即所在集合的根节点又当如何，将其删除的话其子节点需要重新建立关系，这个过程还是有一定复杂性的。我们可以采取另外的思路：二次哈希法（ReHash），对于每个结点都有一个哈希值，在进行查找之前需要将x转化成它的哈希值HASH[x]，那么在进行删除的时候，只要将x的哈希值进行改变，变成一个从来没有出现过的值（可以采用一个计数器来实现这一步），然后对新的值建立集合，因为只有它一个元素，所以必定是一个新的集合。这样做可以保证每次删除操作的时间复杂度都是O(1)的，而且不会破坏原有树结构，唯一的一个缺点就是每删除一个结点其实是多申请了一块内存，如果删除操作无限制，那么内存会无限增长。

简单的说，就是建立虚拟节点，这样原来的节点还在树当中，并不会破坏其结构，开辟一个id[ ]数组表示元素 x 对应的值。

#include<bits/stdc++.h>using namespace std;const int maxn = 100005;int cnt,fa[maxn],num[maxn],sum[maxn],id[maxn];void init(int N){	for (int i = 0;i <= N;i++)	{		fa[i] = id[i] = sum[i] = i;		num[i] = 1;	}	cnt = N;}int find(int x){	int r = x;	while (r != fa[r])	r = fa[r];	int i = x,j;	while (i != r)	{		j = fa[i];		fa[i] = r;		i = j;	}	return r;}void Union(int x,int y){	int fx = find(id[x]),fy = find(id[y]);	fa[fx] = fy;	num[fy] += num[fx];	sum[fy] += sum[fx];}void Delete(int x){	int fx = find(id[x]);	--num[fx];	sum[fx] -= x;	id[x] = ++cnt,fa[id[x]] = id[x],num[id[x]] = 1,sum[id[x]] = x;}int main(){	int N,M;	while (~scanf("%d%d",&N,&M))	{		int opt,x,y;		init(N);		while (M--)		{			scanf("%d",&opt);			if (opt == 1)			{				scanf("%d%d",&x,&y);				if (find(id[x]) != find(id[y]))	Union(x,y);			}			else if (opt == 2)			{				scanf("%d%d",&x,&y);				if (find(id[x]) != find(id[y]))	Delete(x),Union(x,y);			}			else			{				scanf("%d",&x);				printf("%d %d\n",num[find(id[x])],sum[find(id[x])]);			}		}	}	return 0;}