Almost Union-Find UVA - 11987（并查集的删除操作）

题意：求出每个集合的元素个数，及总和，给出三个操作：

1 将含有a元素和b元素的集合合并；2 将a元素放入含有b元素的集合中；3 输出a元素所在集合的元素个数及总和；

思路：正常并查集，与并查集元素的删除

题目：

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 operati

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}.

Input

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).

Output

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}

Sample Input

5 7

1 1 2

2 3 4

1 3 5

3 4

2 4 1

3 4

3 3

Sample Output

3 12

3 7

2 8

/*对于删除操作，在完美的并查集中（所有节点都直接连接在根节点上），理论上只要把要删除的节点的上级重新指向自己就可以了。
但是实际情况中，我们的并查集形成的树的形态都是不可预估形态的，如果直接将一个节点指向自己可能会将他的“下级”和他一起删除，这就和我们的想法违背了。
所以在一个需要删除的并查集中初始化时就要处理一下：
首先可以将每一个点都设立一个虚拟父节点，这样根节点就是我们设立的虚拟节点，类似于将每个节点放到一个盒子中
如果删除某点，那么可以修改当前节点的父节点来导致当前点的孤立，即删除时把这个节点从当前盒子拿出来，放到另一个盒子中。
由于节点之间都是通过盒子来确定关系的，所以盒子中元素是否存在并不影响节点之间的关系。*/
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<queue>
#define Lint long long int
using namespace std;
const int MAXN=200010;
int f[MAXN];/*盒子内的元素链接*/
int sum[MAXN];/*集合内元素之和*/
int p[MAXN];/*盒子*/
int siz[MAXN];/*集合内元素个数，下标为盒子下标*/
int n,m,cnt;
int find(int x)
{
    return x==f[x] ? f[x] : f[x]=find( f[x] ) ;
}
int main()
{
    int opt,u,v,x,y;
    while( scanf("%d%d",&n,&m)!=EOF )//删除节点，就是把原先的节点设置为虚点，然后把点的位置用num数组指向新的位置。
    {
        cnt=n;
        for(int i=1; i<=n; i++)   f[i]=p[i]=sum[i]=i,siz[i]=1;
        for(int i=1; i<=m; i++)
        {
            scanf("%d",&opt);
            if( opt==1 )
            {
                scanf("%d%d",&u,&v);
                u=p[u],v=p[v];
                u=find( u ),v=find( v );
                if( u==v )   continue ;
                f[u]=v;
                siz[v]+=siz[u],sum[v]+=sum[u];
            }
            if( opt==2 )
            {
                scanf("%d%d",&u,&v);
                x=find( p[u] ),y=find( p[v] );
                if( x==y )   continue ;
                sum[x]-=u,siz[x]--;/*盒子的名称不变，除去该元素*/
                x=p[u]=++cnt;/*重新申请一个内存，里面只有要操作的元素，改变该元素的祖先*/
                f[x]=y;
                sum[y]+=u,siz[y]++;
            }
            if( opt==3 )
            {
                scanf("%d",&u);
                u=find( p[u] );
                printf("%d %d\n",siz[u],sum[u]);
            }
        }
    }
    return 0;
}