Almost Union-Find（并查集 删除）——UVa 11987

题目：

Almost Union-Find
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 4
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}

 

 

题意：

给出n个节点，有三种操作，

1 ：将两个节点所在集合合并；

2：将一个元素从一个集合移动到另一个集合；

3：查询指定元素所在集合中的元素个数，和权值总和；

 

思路：

开一个sum数组和num数组，分别记录集合中的权值总和，和元素个数，

合并的时候利用并查集直接合并，但是操作二移动的时候就会出现问题，

如果移动的是集合中的根节点时，直接处理会把整个集合移动到另一个集合中去，

所以要再开一个数组 id 记录每个节点的真实下标，每次要移动时，

先消除这个元素在原集合中的影响，再新开一个点来表示这个元素，再把这个点

与集合合并。

 

代码：

 

#include<bits/stdc++.h>
#define ll long long
#define N 1008611
#define inf 0x3f3f3f3f
using namespace std;
int f[200010],sum[200010],num[200010],id[200010];
int getf(int x)
{
    if(x==f[x])
        return x;
    return f[x]=getf(f[x]);
}
void Inint()
{
    for(int i=0; i<100010; i++)
    {
        f[i]=i;
        sum[i]=i;
        id[i]=i;
        num[i]=1;
    }
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        int nn=n+1;
        Inint();
        for(int i=0; i<m; i++)
        {
            int x;
            scanf("%d",&x);
            if(x==1)//合并集合
            {
                int p,q;
                scanf("%d%d",&p,&q);
                int t1=id[p];//找到p q 真实的表示
                int t2=id[q];
                t1=getf(t1);
                t2=getf(t2);
                if(t1==t2)
                    continue;
                f[t2]=t1;
                sum[t1]+=sum[t2];
                num[t1]+=num[t2];
            }
            else if(x==2)//将元素从一个集合移动到另一个集合中
            {
                int p,q;
                scanf("%d%d",&p,&q);
                int t1=id[p];
                int t2=id[q];
                t1=getf(t1);
                t2=getf(t2);
                if(t1==t2)
                    continue;
                sum[t1]-=p;//消除 p 在原来集合中的影响
                num[t1]-=1;
                id[p]=nn++;//新建一个节点来表示 p
                t1=id[p];
                f[t1]=t2;
                sum[t2]+=p;//移动到另一个集合
                num[t2]+=1;
            }
            else if(x==3)
            {
                int y;
                scanf("%d",&y);
                y=getf(id[y]);
                printf("%d %d\n",num[y],sum[y]);
            }
        }
    }
    return 0;
}