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6012: Shave Beaver!

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题目描述

B2. Shave Beaver!
B2.剃海狸！
智能海狸最近设计并制造了一种创新的纳米技术通用海狸大规模剃须机“海狸5000”。海狸剃5000只可以按家庭剃海狸！它是如何工作的？非常容易！
有n个海狸，每个海狸都有一个从1到n的唯一id。考虑一个排列a1，a2，。。。，这些海狸中的一只。海狸剃5000需要一个会话才能剃掉ID从x到y（含）的海狸，如果且仅当存在这样的索引i1＜i2＜<即ai1=x，ai2=x+1。。。，aik-1＝y-1，aik＝y。这真的很方便。例如，它需要一个会话来剃掉海狸1、2、3、…的排列，。。。，n
如果我们不能在一个会话中将海狸从x剃到y，那么我们可以将这些海狸分成组[x，p1]，[p1+1，p2]。。。，[pm+1，y]（x≤p1<p2<…<pm<y），这样机器可以在一个会话中为每组的海狸剃毛。但是，Beaversaw 5000需要m+1个工作会话才能将海狸从x剃到y。
所有的海狸都不安分，它们一直试图交换。因此，如果我们更正式地考虑这个问题，我们可以考虑两种类型的查询：
Beaversaw 5000为ID从x到y（包括x和y）的海狸剃毛所需的最少会话数是多少？
x和y位置的两只海狸（海狸ax和ay）互换了位置。
你可以假设任何海狸都可以剃任何次数。

要获得30分，需要解决约束条件：n≤100（子问题B1）；

要获得100分，您需要解决具有约束的问题：n≤3·105（子问题B1+B2）。

注意，在子问题B1和子问题B2中，查询的数量q都被限制为1≤q≤105。

The Smart Beaver has recently designed and built an innovative nanotechnologic all-purpose beaver mass shaving machine, "Beavershave 5000". Beavershave 5000 can shave beavers by families! How does it work? Very easily!
There are n beavers, each of them has a unique id from 1 to n. Consider a permutation a1,a2,...,an of n these beavers. Beavershave 5000 needs one session to shave beavers with ids from x to y (inclusive) if and only if there are such indices i1<i2<...<ik, that ai1=x, ai2=x+1, ..., aik-1=y-1, aik=y. And that is really convenient. For example, it needs one session to shave a permutation of beavers 1,2,3,...,n.
If we can't shave beavers from x to y in one session, then we can split these beavers into groups [x,p1], [p1+1,p2], ..., [pm+1,y] (x≤p1<p2<...<pm<y), in such a way that the machine can shave beavers in each group in one session. But then Beavershave 5000 needs m+1 working sessions to shave beavers from x to y.
All beavers are restless and they keep trying to swap. So if we consider the problem more formally, we can consider queries of two types:
  • what is the minimum number of sessions that Beavershave 5000 needs to shave beavers with ids from x to y, inclusive?
  • two beavers on positions x and y (the beavers ax and ay) swapped.
You can assume that any beaver can be shaved any number of times.
Input
The first line contains integer n − the total number of beavers, 2≤n. The second line contains n space-separated integers − the initial beaver permutation.
The third line contains integer q − the number of queries, 1≤q≤105. The next q lines contain the queries. Each query i looks as pi xi yi, where pi is the query type (1 is to shave beavers from xi to yi, inclusive, 2 is to swap beavers on positions xi and yi). All queries meet the condition: 1≤xi<yi≤n.
  • to get 30 points, you need to solve the problem with constraints: n≤100 (subproblem B1);
  • to get 100 points, you need to solve the problem with constraints: n≤3·105 (subproblems B1+B2).
Note that the number of queries q is limited 1≤q≤105 in both subproblem B1 and subproblem B2.
Output
For each query with pi=1, print the minimum number of Beavershave 5000 sessions.
Examples
Input 
5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5
Output 
2
1
3
5

输入格式

第一行包含整数n−海狸总数，2≤n。第二行包含n个空间分隔的整数——初始海狸排列。

第三行包含整数q−查询数，1≤q≤105。接下来的q行包含查询。每个查询我看起来都是pi xi yi，其中pi是查询类型（1是将海狸从xi刮到yi，包括xi和yi），2是在位置xi和yi上交换海狸）。所有查询都满足条件：1≤xi<yi≤n。

输出格式

对于pi=1的每个查询，打印Beaversaw 5000个会话的最小数量。

输入样例 复制 

5
1 3 4 2 5
6
1 1 5
1 3 4
2 2 3
1 1 5
2 1 5
1 1 5

输出样例 复制 

2
1
3
5

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