CDQ(动态逆序对的几种做法)

菜鸡总觉得自己会了很多东西 然而在学长的鞭策下 还有很多需要加深的  希望再一次突破自己的数据结构！

#include <bits/stdc++.h>
#define f first
#define s second
#define ll long long
#define pii pair<ll,int>
#define inc(i,l,r) for(int i=l;i<=r;i++)
const int MAXN=1e5+10;
using namespace std;
ll read(){  
    ll x=0,f=1;char ch=getchar();  
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}  
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();  
    return f*x;
}
int a[MAXN];int b[MAXN],c[MAXN];
int G[MAXN];
ll ans[MAXN];ll d[MAXN];
int n,m;
int get_id(int x){return x&(-x);}
ll Sum(int x){
	ll sum=0;
	for(int i=x;i>0;i-=get_id(i))sum+=d[i];
	return sum;
}
void update(int x,int vul){
	for(int i=x;i<=m+1;i+=get_id(i))d[i]+=vul;
}
void update1(int x,int vul){
	for(int i=x;i<=n;i+=get_id(i))d[i]+=vul;
}
void clean(int x){
	for(int i=x;i<=m+1;i+=get_id(i))d[i]=0;
}
void querty(int l,int r,int mid){
	int i=l,j=mid+1;G[0]=0;
	while(i<=mid&&j<=r){
		while(i<=mid&&c[i]<c[j]){
			G[++G[0]]=c[i];
			update(b[c[i]],1);i++;
		}
		ans[b[c[j]]]+=Sum(m+1)-Sum(b[c[j]]);
		G[++G[0]]=c[j];
		j++;
	}
	if(i<=mid){
		for(;i<=mid;i++)G[++G[0]]=c[i];
	}
	if(j<=r){
		for(;j<=r;j++){
			ans[b[c[j]]]+=Sum(m+1)-Sum(b[c[j]]);
			G[++G[0]]=c[j];
		}
	}
	for(int i=1;i<=G[0];i++)c[l+i-1]=G[i],clean(b[G[i]]);
}
void cdq(int l,int r){
	if(l>=r)return ;
	int mid=(l+r)>>1;
	cdq(l,mid);
	cdq(mid+1,r);
	querty(l,r,mid);
}
void querty1(int l,int r,int mid){
	int i=l,j=mid+1;G[0]=0;
	while(i<=mid&&j<=r){
		while(i<=mid&&a[i]>a[j]){
			G[++G[0]]=a[i];
			update(b[a[i]],1);i++;
		}
		ans[b[a[j]]]+=Sum(m+1)-Sum(b[a[j]]);
		G[++G[0]]=a[j];
		j++;
	}
	if(i<=mid){
		for(;i<=mid;i++)G[++G[0]]=a[i];
	}
	if(j<=r){
		for(;j<=r;j++){
			ans[b[a[j]]]+=Sum(m+1)-Sum(b[a[j]]);
			G[++G[0]]=a[j];
		}
	}
	for(int i=1;i<=G[0];i++)a[l+i-1]=G[i],clean(b[G[i]]);
}
void cdq1(int l,int r){
	if(l>=r)return ;
	int mid=(l+r)>>1;
	cdq1(l,mid);
	cdq1(mid+1,r);
	querty1(l,r,mid);
}
int main(){
	n=read();m=read();
	inc(i,1,n)a[i]=read(),c[n-i+1]=a[i];
	int vul;
	inc(i,1,m)vul=read(),b[vul]=i;
	inc(i,1,n)if(!b[i])b[i]=m+1;
	ll sum=0;
	for(int i=1;i<=n;i++){
		sum+=(Sum(n)-Sum(a[i]));
		update1(a[i],1);
	}
	for(int i=1;i<=n;i++)clean(a[i]);
	cdq(1,n);
	cdq1(1,n);
	printf("%lld\n",sum);
	for(int i=1;i<m;i++)sum-=ans[i],printf("%lld\n",sum);
	return 0;
}