B2. Shave Beaver!
 

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, thatai1 = xai2 = 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 xiyi, 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

 题意:

  给你长度n的序列,m次询问

  1:x -> y 的花费  满足 每次 选择 以一个a值  能到大其右边任意位置 (即最长连续上升子序列)算一次路径,问从x值到达y值,需要几次

  2:x,y  交换a[x],a[y];

题解:

  假设x+1在 x的右边 那么此x的位置值为 1,即任意的区间求和

  有交换操作,线段树维护a[x],a[y]对序列的影响即可

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 3e5+, M = 2e5++, mod = 1e9+, inf = 0x3fffffff; int id[N],a[N],n,m,v[N*];
void update(int i,int ll,int rr,int x,int c) {
if(ll == rr) {
v[i] = c;
return ;
}
if(x <= mid) update(ls,ll,mid,x,c);
else update(rs,mid+,rr,x,c);
v[i] = v[ls] + v[rs];
}
int ask(int i,int ll,int rr,int x,int y) {
if(ll == x && y == rr) {
return v[i];
}
if(y <= mid) return ask(ls,ll,mid,x,y);
else if(x > mid) return ask(rs,mid+,rr,x,y);
else return ask(ls,ll,mid,x,mid) + ask(rs,mid+,rr,mid+,y);
}
int main() {
scanf("%d",&n);
for(int i = ; i <= n; ++i) scanf("%d",&a[i]),id[a[i]] = i;
for(int i = ; i < n; ++i) {
if(id[i] > id[i+]) update(,,n,i,);
}
scanf("%d",&m);
for(int i = ; i <= m; ++i) {
int op,x,y;
scanf("%d%d%d",&op,&x,&y);
if(op == ) {
printf("%d\n",ask(,,n,x,y-) + );
} else {
int tmp1 = a[x];
int tmp2 = a[y];
int tt = id[a[x]];
id[a[x]] = id[a[y]];
id[a[y]] = tt;
swap(a[x],a[y]);
if(tmp1+ <= n && id[tmp1] > id[tmp1+]) update(,,n,tmp1,);
if(tmp1- >= && id[tmp1-] < id[tmp1]) update(,,n,tmp1-,); if(tmp2+ <= n && id[tmp2] < id[tmp2+]) update(,,n,tmp2,);
if(tmp2- >= && id[tmp2-] > id[tmp2]) update(,,n,tmp2-,);
}
}
return ;
}

  

codeforce ABBYY Cup 3.0 - Finals (online version) B2. Shave Beaver! 线段树的更多相关文章

  1. ABBYY Cup 3.0 - Finals (online version)

    A 开个数组记录一下 #include <iostream> #include<cstdio> #include<cstring> #include<algo ...

  2. Codeforces Round #535 (Div. 3) E2. Array and Segments (Hard version) 【区间更新 线段树】

    传送门:http://codeforces.com/contest/1108/problem/E2 E2. Array and Segments (Hard version) time limit p ...

  3. VK Cup 2015 - Qualification Round 1 D. Closest Equals 离线+线段树

    题目链接: http://codeforces.com/problemset/problem/522/D D. Closest Equals time limit per test3 secondsm ...

  4. C.Fountains(Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)+线段树+RMQ)

    题目链接:http://codeforces.com/contest/799/problem/C 题目: 题意: 给你n种喷泉的价格和漂亮值,这n种喷泉题目指定用钻石或现金支付(分别用D和C表示),C ...

  5. Codeforces VK Cup 2015 - Qualification Round 1 D. Closest Equals 离线线段树 求区间相同数的最小距离

    D. Closest Equals Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline/prob ...

  6. Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals) Problem E (Codeforces 831E) - 线段树 - 树状数组

    Vasily has a deck of cards consisting of n cards. There is an integer on each of the cards, this int ...

  7. composer install 遇到问题 Problem 1 - phpunit/phpunit 5.7.5 requires php ^5.6 || ^7.0 -> your PHP version (5.5.3 0) does not satisfy that requirement.

    $ composer install Loading composer repositories with package information Updating dependencies (inc ...

  8. Pycharm 中You are using pip version 10.0.1, however version 18.1 is available. You should consider upgrading via the 'python -m pip install --upgrade pip' command.

    今天运行程序的时候出现了: You are using pip version 10.0.1, however version 18.1 is available.You should conside ...

  9. pip install psutil出错-You are using pip version 10.0.1, however version 18.0 is available.

    今天想用python代替shell做运维相关的事,写代码都是在本机,调试在服务器上 C:\Users\0>pip install psutilRequirement already satisf ...

随机推荐

  1. WinForm开发框架--动态读取DLL模式

    1\ WinForm开发框架--动态读取DLL模式   http://www.2cto.com/kf/201306/217199.html 2\ 广州爱奇迪     http://www.iqidi. ...

  2. 设计算法,求AB两个整数集合的交集

    [本文链接] http://www.cnblogs.com/hellogiser/p/ab-set-intersection.html [分析] 思路1:排序法 对集合A和集合B进行排序(升序,用快排 ...

  3. Qt字符转换

    1.QString  -> char* #include<QTextCodec> QTextCodec::setCodecForLocale(QTextCodec::codecFor ...

  4. Python~if,while,for~顺序,判断,循环

    if A: for -in : while x: if A:elif:else:       不能直接用int进行迭代,而必须加个range.     range(len(L))     int ob ...

  5. 4.kvm克隆虚拟机

    virt-clone 作用简介 virt-clone 主要是用来克隆kvm虚拟机,并且通过 Options.General Option.Storage Configuration.Networkin ...

  6. 零件分组_DP

    问题 C: 零件分组 时间限制: 1 Sec  内存限制: 64 MB提交: 31  解决: 14[提交][状态][讨论版] 题目描述 某工厂生产一批棍状零件,每个零件都有一定的长度(Li)和重量(W ...

  7. js中 map 遍历数组

    map 方法会迭代数组中的每一个元素,并根据回调函数来处理每一个元素,最后返回一个新数组.注意,这个方法不会改变原始数组. 在我们的例子中,回调函数只有一个参数,即数组中元素的值 (val 参数) , ...

  8. C#一维数组

    数组:相同数据类型的元素按照一定的顺序进行排列生成的集合(一组数据)一维数组:int [] array=new int[5];int[] array = new int[] {1,2,3,4,5 }; ...

  9. 【xml】转义字符 &lt;等符号出现的原因

    来源:http://www.cnblogs.com/hyd309/p/3549076.html HTML中的转义字符  HTML中<, >,&等有特别含义,(前两个字符用于链接签, ...

  10. 20145213《Java程序设计》第十周学习总结

    20145213<Java程序设计>第十周学习总结 教材学习总结 网络编程 网络编程就是在两个或两个以上的设备(例如计算机)之间传输数据.程序员所作的事情就是把数据发送到指定的位置,或者接 ...