UVA-11987-Almost Union-Find（并查集删除）

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}.

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). The size of input file does not exceed 5MB.

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

Sample Input
5 7
1 1 2
2 3 4
1 3 5
3 4
2 4 1
3 4x
3 3
Output for the Sample Input
3 12
3 7
2 8
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}

思路：并查集删除。被删除的节点的关系不能在原集合里删除，因为其子节点需要它来和爷爷节点联系。通过建立虚拟节点。

在没删除之前的操作就用id[x]代替x。如果有删除操作，之后对x的操作就是在对id[x]进行操作。

PS：真正的并查集删除操作是不存在的，建立虚拟编号相当于克隆了一个“被删除”的节点，原来的集合里还是存在“被删除”的节点的。只是以后只调用编号了而不会调用真正的节点本身。

#include<cstdio>
using namespace std;
int pre[200005],num[200005],sum[200005],id[100005];//id为虚拟编号节点 
int n,m,l;

void init(){
    l=n;
    for(int i=0;i<=n;i++){
        sum[i]=pre[i]=id[i]=i;
        num[i]=1;        
    }
} 

int find(int x){
    while(pre[x]!=x){
        int r=pre[x];
        pre[x]=pre[r];
        x=r;
    }
    return x;
}

void merge(int x,int y){
    int fx=find(x);
    int fy=find(y);
    if(fx!=fy)     {
    pre[fx]=fy;
    num[fy]+=num[fx];
    sum[fy]+=sum[fx];
    }
}

void move(int x){
    int fx=find(id[x]);// 移走后id[x]被更新，但pre[x]不能被更新，要保留，所以只能用虚拟节点! 
    sum[fx]-=x;            
    num[fx]--;             
    id[x]=++l; //x是第l-n个被删除的节点 
    sum[id[x]]=x;
    num[id[x]]=1;
    pre[id[x]]=id[x];
}

int main(){
    int d,p,q;
    while(~scanf("%d%d",&n,&m)){
        init();    
        while(m--){
            scanf("%d%d",&d,&p);
            if(d==1){
                scanf("%d",&q);
                merge(id[p],id[q]);
            }
             else if(d==2){
                scanf("%d",&q);
                if(find(id[p])!=find(id[q])){            
                    move(p); 
                    merge(id[p],id[q]);        
                 }
            }
            else{
                int fp=find(id[p]);
                printf("%d %d
",num[fp],sum[fp]);
            }
        }    
    }
    return 0;
}