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 = 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.
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.
For each query with pi = 1, print the minimum number of Beavershave 5000 sessions.
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
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+10, M = 2e5+35000+11, mod = 1e9+7, inf = 0x3fffffff; int id[N],a[N],n,m,v[N*20]; 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+1,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+1,rr,x,y); else return ask(ls,ll,mid,x,mid) + ask(rs,mid+1,rr,mid+1,y); } int main() { scanf("%d",&n); for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),id[a[i]] = i; for(int i = 1; i < n; ++i) { if(id[i] > id[i+1]) update(1,1,n,i,1); } scanf("%d",&m); for(int i = 1; i <= m; ++i) { int op,x,y; scanf("%d%d%d",&op,&x,&y); if(op == 1) { printf("%d ",ask(1,1,n,x,y-1) + 1); } 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+1 <= n && id[tmp1] > id[tmp1+1]) update(1,1,n,tmp1,1); if(tmp1-1 >= 1 && id[tmp1-1] < id[tmp1]) update(1,1,n,tmp1-1,0); if(tmp2+1 <= n && id[tmp2] < id[tmp2+1]) update(1,1,n,tmp2,0); if(tmp2-1 >= 1 && id[tmp2-1] > id[tmp2]) update(1,1,n,tmp2-1,1); } } return 0; }