UVA

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).
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

题目意思：

初始时，一共有n个元素的组合。

给出三个操作：（k p q）

k==1 p q：合并p, q所在的集合

k==2 p q：把p移动到q所在的集合

k==3 p：输出p所在的集合的元素的个数和元素之和

分析：

虚结点思想

并查集中并没有删除的操作。
所以就需要将删除的这个点的影响降到0，
也就是给删除的点申请一个新的id，
以后都是用这个新的id来表示这个点，
这样原来的那个集合里的点p就没意义，自然影响就为0.

 

#include <iostream>
#include<algorithm>
#include <cstdio>
#include<cstring>
#include<math.h>
#include<memory>
using namespace std;
typedef long long LL;
#define max_v 200005
int pa[max_v];
int sum[max_v];//sum[x] x（根结点）所在元素集合所有元素的和
int id[max_v];//id[x]=y 表示x结点现在的编号为y
int cnt[max_v];//cnt[x] x(根结点）所在集合元素的个数

int n,m,tot;

void make_set(int x)
{
    pa[x]=sum[x]=id[x]=x;
    cnt[x]=1;
}
int find_set(int x)
{
    if(x!=pa[x])
        pa[x]=find_set(pa[x]);
    return pa[x];
}
void union_set(int x,int y)//x合并到y
{
    x=find_set(x);
    y=find_set(y);
    if(x==y)
        return ;

    pa[x]=y;
    sum[y]+=sum[x];
    cnt[y]+=cnt[x];

}
void del(int x)
{
    int fx=find_set(id[x]);
    sum[fx]-=x;
    cnt[fx]--;

    tot++;
    id[x]=tot;
    pa[tot]=tot;
    sum[tot]=x;
    cnt[tot]=1;
}
int main()
{
    int n,m,k;
    int x,y;
    while(~scanf("%d %d",&n,&m))
    {
        tot=n;
        for(int i=0;i<=n;i++)
            make_set(i);
        for(int i=0;i<m;i++)
        {
            scanf("%d",&k);
            if(k==1)//x合并到y
            {
                scanf("%d %d",&x,&y);
                union_set(id[x],id[y]);
            }else if(k==2)//x从原有集合拿出放入到y
            {
                scanf("%d %d",&x,&y);
                int fx=find_set(id[x]);
                int fy=find_set(id[y]);

                if(fx!=fy)
                {
                    del(x);
                    union_set(id[x],id[y]);
                }
            }else//输出x所在集合元素个数和所有元素和
            {
                scanf("%d",&x);
                int fx=find_set(id[x]);
                printf("%d %d
",cnt[fx],sum[fx]);
            }
        }
    }
    return 0;
}