牛客,似花还似非花(dp优化)

似花还似非花
这个题实际上是求解两个排列的包含某一特定元素的情况下的最长公共子序列长度。
最长上升子序列用树状数组优化

//#pragma GCC optimize(2)
//#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define yes cout<<"Yes\n"
#define no cout<<"No\n"
#define yesno if(fg)yes;else no;
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
const int inf=8e18;
const int maxn=2e5+100;
int a[maxn],b[maxn],c[maxn],c2[maxn],pos[maxn];
int dp[maxn],dp2[maxn];
int lowbit(int x)
{
	return x&(-x);
}
void update(int x,int val)
{
	while(x<maxn)
	{
		c[x]=max(c[x],val);
		x+=lowbit(x);
	}
}
void update2(int x,int val)
{
	while(x>0)
	{
		c2[x]=max(c2[x],val);
		x-=lowbit(x);
	}
}
int ask(int x)
{
	int ans=0;
	while(x>0)
	{
		ans=max(ans,c[x]);
		x-=lowbit(x);
	}
	return ans;
}
int ask2(int x)
{
	int ans=0;
	while(x<maxn)
	{
		ans=max(ans,c2[x]);
		x+=lowbit(x);
	}
	return ans;
}
signed main()
{
	IOS
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x;
		cin>>x;
		pos[x]=i;
	}
	for(int i=1;i<=n;i++)
	{
		int x;
		cin>>x;
		a[i]=pos[x];
		b[x]=i;
	}
	for(int i=1;i<=n;i++)
	{
		dp[i]=ask(a[i])+1;
		update(a[i],dp[i]);
	}
	for(int i=n;i>=1;i--)
	{
		dp2[i]=ask2(a[i])+1;
		update2(a[i],dp2[i]);
	}

	int q;
	cin>>q;
	while(q--)
	{
		int x;
		cin>>x;
		x=b[x];
		cout<<n-dp[x]-dp2[x]+1<<"\n";
	}
}