Description
Input
Output
Sample Input
1
5
3
4
2
5
1
4
2
Sample Output
2
2
1
樣例解釋
(1,5,3,4,2)(1,3,4,2)(3,4,2)(3,2)(3)。
HINT
N<=100000 M<=50000
正解:CDQ分治。
這題用來考試,一堆50分暴力,一人寫出正解但是沒開long long。。
考慮把刪除變成插入,那麼每次插入是按照時間排序的。那麼只要滿足i<j,ai>aj,ti<tj,那麼這就是一個逆序對。於是這題就變成裸的三維偏序了。
//It is made by wfj_2048~
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#define inf (1<<30)
#define il inline
#define RG register
#define ll long long
#define lb(x) (x & -x)
#define File(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
using namespace std;
struct node{ int x,y,t; }q[100010],qu[100010];
ll c[100010],ans[100010],Ans;
int match[100010],n,m;
il int gi(){
RG int x=0,q=0; RG char ch=getchar(); while ((ch<'0' || ch>'9') && ch!='-') ch=getchar();
if (ch=='-') q=1,ch=getchar(); while (ch>='0' && ch<='9') x=x*10+ch-48,ch=getchar(); return q ? -x : x;
}
il int cmp(const node &a,const node &b){ return a.x<b.x || (a.x==b.x && a.y<b.y) || (a.x==b.x && a.y==b.y && a.t<b.t); }
il void add(RG int x,RG int v){ for (RG int i=x;i<=n;i+=lb(i)) c[i]+=(ll)v; return; }
il ll query(RG int x){ RG ll res=0; for (RG int i=x;i;i-=lb(i)) res+=c[i]; return res; }
il void solve(RG int l,RG int r){
if (l>=r) return; RG int mid=(l+r)>>1,t1=l-1,t2=mid;
for (RG int i=l;i<=r;++i) if (q[i].t<=mid) add(q[i].y,1); else ans[q[i].t]+=query(n)-query(q[i].y);
for (RG int i=l;i<=r;++i) if (q[i].t<=mid) add(q[i].y,-1);
for (RG int i=r;i>=l;--i) if (q[i].t<=mid) add(q[i].y,1); else ans[q[i].t]+=query(q[i].y);
for (RG int i=r;i>=l;--i) if (q[i].t<=mid) add(q[i].y,-1);
for (RG int i=l;i<=r;++i) if (q[i].t<=mid) qu[++t1]=q[i]; else qu[++t2]=q[i];
for (RG int i=l;i<=r;++i) q[i]=qu[i]; solve(l,mid),solve(mid+1,r); return;
}
il void work(){
n=gi(),m=gi(); for (RG int i=1;i<=n;++i) q[i].x=i,q[i].y=gi(),match[q[i].y]=i; RG int ti=n,v;
for (RG int i=1;i<=m;++i) v=gi(),q[match[v]].t=ti--; for (RG int i=1;i<=n;++i) if (!q[i].t) q[i].t=ti--;
sort(q+1,q+n+1,cmp); solve(1,n); for (RG int i=1;i<=n;++i) Ans+=ans[i];
for (RG int i=n;i>n-m;--i){ printf("%lld\n",Ans); Ans-=ans[i]; } return;
}
int main(){
File("dynamic");
work();
return 0;
}