UVA11987(带权并查集的删除操作)

Almost Union-Find

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=57812#problem/C

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}

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Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!

解题思路：

一道加权并查集的题，但在这基础上，加入了并查集的删除操作。题意是说n个节点进行m次操作，如果为1操作，

那么把a和b两个操作连接；如果为2操作，那么把a节点全部移到b上，即认b为祖先；如果为3，那么询问a节点所

属的树有多少结点，该树所有结点元素的总和为多少。i 节点的元素初值为 i 。

大体思路很像昨晚做的加权并查集。开sum记录该节点所在集合的元素总值，开cnt记录该节点所在集合有多少元

素。对于本道题涞说，难点是当执行2操作时，涉及到一个a节点连到b节点上的操作，这是要在a原来所属集合中删除a，

并在b集合中新增a，相应的cnt 和sum值都要做修改。

我们额外再开一个id数组，记录当前节点的编号，当执行2  a b ， 把a的元素放到b上之前，把a做下处理。有种思

想是直接把a指向b，如果a是当前集合的叶子节点，那么这种思想是成立的，可是如果a是当前集合的非叶子节点，那么

当a指向b后必然会导致a的孩子也指向了b，所以这种思想是有局限性的。

我们可以额外再开一个节点记录a的信息，然后把a当前集合做处理（sum - a , cnt - 1），相当于把该集合的a节点滞

空。然后把新建立的节点连到b上，再对b进行处理（sum + sum[a] , cnt + 1）。

最后在询问时只要找到祖先，然后输出一下就好了。sum开成long long ，因为n给的1e5······微大······

完整代码：

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