第十四届华科校赛 C-Professional Manager(并查集的拆分)

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
 64bit IO Format: %lld

题目描述： https://www.nowcoder.com/acm/contest/106/C

     It’s universally acknowledged that there’re innumerable trees in the campus of HUST.               Thus a professional tree manager is needed. Your task is to write a program to help manage the trees.        Initially, there are n forests and for the i-th forest there is only the i-th tree in it. Given four kinds of operations.      1 u v, merge the forest containing the u-th tree and the forest containing the v-th tree;    2 u, separate the u-th tree from its forest;    3 u, query the size of the forest which contains the u-th tree;    4 u v, query whether the u-th tree and the v-th tree are in the same forest.
输入描述:
The first line contains an integer T, indicating the number of testcases.In each test case:The first line contains two integers N and Q, indicating the number of initial forests and the number of operations.Then Q lines follow, and each line describes an operation.
输出描述:
For each test cases, the first line should be "Case #i:", where i indicate the test case i.For each query 3, print a integer in a single line.For each query 4, print "YES" or "NO" in a single line.

示例1

输入

1
10 8
3 1
4 1 2
1 1 2
3 1
4 1 2
2 1
3 1
4 1 2

输出

Case #1:
1
NO
2
YES
1
NO

题目大意：有n棵树，编号为1~n，接下来又4种操作：1、合并树u和树v为一片森林。2、把树u和原来的森林分开。3、查询树u所在的森林的树的总数。4、查询树u和数v是否在一片森林。

解题思路：从操作来看很容易想到并查集的做法，但是对于操作2，我们似乎感觉到不太好实现，如果不是根节点，我们就直接root[a] = a就可以了，然而根节点本来就是指向它自己，要是根节点也有子节点的性质就好了，所以我们不妨给每个树开始都创建一个根节点，这个虚构的根节点不计入总数即可，这样我们就很轻松的解决了并查集的拆分问题。

AC代码：

#include
#include
#include
#include
#include
#include
#include
#include