E. Intersection of Permutations

题意:给两个排列,2种操作1,查询两个区间a和b一样的值个数,2,交换b的两个值
题解:树套树,先把a变成1到n的排列,对b做相同的变换,然后问题就变成了查询区间lb,rb中la到ra的个数,带修改可以树状数组套主席树,需要内存回收

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
//#pragma GCC optimize(4)
//#pragma GCC optimize("unroll-loops")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include<bits/stdc++.h>
#define fi first
#define se second
#define db double
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 998244353
#define ld long double
//#define C 0.5772156649
//#define ls l,m,rt<<1
//#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define ull unsigned long long
//#define base 1000000000000000000
#define fin freopen("a.txt","r",stdin)
#define fout freopen("a.txt","w",stdout)
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
template<typename T>inline T const& MAX(T const &a,T const &b){return a>b?a:b;}
template<typename T>inline T const& MIN(T const &a,T const &b){return a<b?a:b;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const ull ba=233;
const db eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=200000+10,maxn=200000+10,inf=0x3f3f3f3f;

int n,m;
int q[N*150],top;
struct bit_seg{
    int root[N],ls[N*150],rs[N*150],sum[N*150],res;
    bit_seg(){res=0;}
    void update(int &o,int pos,int v,int l,int r)
    {
        if(!o)
        {
            if(top)o=q[top--];
            else o=++res;
        }
        sum[o]+=v;
        if(l==r)return ;
        int m=(l+r)>>1;
        if(pos<=m)update(ls[o],pos,v,l,m);
        else update(rs[o],pos,v,m+1,r);
        if(!sum[o])q[++top]=o,o=0;
    }
    int query(int L,int R,int o,int l,int r)
    {
        if(!o)return 0;
        if(L<=l&&r<=R)return sum[o];
        int m=(l+r)>>1,ans=0;
        if(L<=m)ans+=query(L,R,ls[o],l,m);
        if(m<R)ans+=query(L,R,rs[o],m+1,r);
        return ans;
    }
    int bitquery(int la,int ra,int lb,int rb)
    {
        int ans=0;
        for(int i=rb;i;i-=i&(-i))ans+=query(la,ra,root[i],1,n);
        for(int i=lb-1;i;i-=i&(-i))ans-=query(la,ra,root[i],1,n);
        return ans;
    }
    void bitupdate(int i,int pos,int v)
    {
        for(;i<=n;i+=i&(-i))update(root[i],pos,v,1,n);
    }
}s;
int a[N],b[N],c[N];
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]),c[a[i]]=i;
    for(int i=1;i<=n;i++)scanf("%d",&b[i]),b[i]=c[b[i]];
    for(int i=1;i<=n;i++)s.bitupdate(i,b[i],1);
    for(int i=1;i<=m;i++)
    {
        int op;scanf("%d",&op);
        if(op==1)
        {
            int la,ra,lb,rb;scanf("%d%d%d%d",&la,&ra,&lb,&rb);
            printf("%d\n",s.bitquery(la,ra,lb,rb));
        }
        else
        {
            int x,y;scanf("%d%d",&x,&y);
            s.bitupdate(x,b[x],-1);s.bitupdate(y,b[y],-1);
            swap(b[x],b[y]);
            s.bitupdate(x,b[x],1),s.bitupdate(y,b[y],1);
        }
    }
    return 0;
}
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