Little Pony and Boast Busters

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Problem Description

"I hereby challenge you, Ponyvillians: anything you can do, I can do better. Any takers? Anyone? Or is Trixie destined to be the greatest equine who has ever lived!?!" — "Boast Busters"

Given two permutation P₀ && P₁ of {0, 1, ..., n - 1}, we define the crossing number of it as follows. Write P₀ from left to right above P₁ and draw a straight line between each same elements. The crossing number of
P₀ and P₁ is the number of pairs of lines that cross.

For example, if n = 5, and P₀ = {0, 1, 2, 3, 4}, and P₁ = {1, 3, 0, 2, 4}, then the crossing number of P₀ and P₁ is 3, as shown in the figure below.

Now given you the two permutation, you need to implement the following operations:

SWAP p a b: swap P_p[a] and P_p[b] (0<=p<=1, 0<=a, b<=n-1).

QUERY: ask the crossing number of the current P₀ and P₁.

Input

Input contains multiple test cases (less than 10). For each test case, the first line contains one integer n (1<=n<=10^5).

The second line contains n integers P₀[0], P₀[1], ..., P₀[n-1].

The third line contains n integers P₁[0], P₁[1], ..., P₁[n-1].

The next line contains one integer q —— the number of operations (1<=q<=10^5). The next q line, each line will contains a operation as we mentioned above.

Output

For each query, output the corresponding result in one line.

Sample Input

5
0 1 2 3 4
1 3 0 2 4
5
QUERY
SWAP 1 2 4
QUERY
SWAP 0 2 4
QUERY

Sample Output

3
6
5

Source

BestCoder Round #7

题意：给出两个0到n-1的排列，然后支持两种操作，QUERY查询两组排列的交叉的元素个数，SWAP 交换一个排列中两个位置的元素

思路：这题最后可以转化为给出一个排列P（将给出的两个排列的位置一一对应起来，参见大神的题解http://bestcoder.hdu.edu.cn），

可以动态交换P的任意两个位置的数，然后查询整个序列的逆序对数

动态逆序对是比较经典的树套树模型，我这里用的是树状数组套treap实现的，之前用的线段树妥妥的超时了

可以先用树状或归并排序求出初始序列的逆序对数，记为ans，然后每次查询只需要输出ans即可

重点在于交换，因为交换以后ans会随之改变，假设现在要交换L，R位置的数，那么[0，L-1]，[R+1,n-1]这些位置的数是不用考虑的，因为它们对中间

元素的逆序对的贡献是不会改变的，所以只用考虑[L+1，R-1]之间的数

假设元素P[L] > P[R] ，设区间[L+1，R-1]的数小于P[L]的个数为L1，大于的个数为L2，同理设小于P[R]的个数为R1，大于的个数为R2，那么交换后的

答案 为ans = ans + L2-L1 + R1-R2 = ans + R1+L2 - (L1+R2)，现在只需求出 R1 + L2 问题即可完美解决，因为L1+R2=2*n-(R1+L2) ，其中n为区间

[L+1，R-1]的元素个数，现在令区间[L+1，R-1]为S

而求R1等价于求S中元素小于P[R]的个数，求L2等价于求S中元素大于P[L]的个数，所以只需要在树状数组里套一个treap，用来维护该树状数组管辖的

区间的元素，具体涉及两种操作：

1.交换元素，相当于先将原位置的数删除，再在新的位置插入，这个就是树状数组单点更新了

2.查询一段区间中所有小于（大于）x的元素个数，相当于区间求和

怎么样，难点全部攻破了吧，这题也是把我写的醉到不行，调了许久才发现竟然树状数组没有初始化，sad~，累觉不爱，可以睡觉去了