bzoj3295(cdq分治)

题目链接

https://www.lydsy.com/JudgeOnline/problem.php?id=3295

题解

说是动态逆序对其实也没那么动态。。如果是删数不如先考虑加数要怎么整。。

加数的话可以把时间作为一维然后就变成了三维偏序问题,这里要统计两个偏序,一个是位置在前的一个是位置在后的,然后直接 $CDQ$ 分治即可




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<cmath>
#include<set>
#include<bitset>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define eps 1e-6
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define NM 100005
#define nm 800005
#define pi 3.1415926535897931
using namespace std;
const ll inf=1e18;
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;
}





struct P{
int id,x,y;ll v;
}a[NM],tmp[NM];
int n,m,tot,c[NM];
ll b[NM];

void add(int x){for(;x<=n;x+=lowbit(x))c[x]++;}
int sum(int x,int s=0){for(;x;x-=lowbit(x))s+=c[x];return s;}
void clr(int x){for(;x<=n&&c[x];x+=lowbit(x))c[x]=0;}

bool cmp(P a,P b){return a.id>b.id;}

void cdq(int l,int r){
if(l==r)return;
int mid=l+r>>1,cnt=l;
cdq(l,mid);cdq(mid+1,r);
for(int x=mid,y=r;x>=l||y>mid;)if(x>=l&&(y<=mid||a[x].x>a[y].x)){
add(a[x].y);x--;
}else{
a[y].v+=sum(a[y].y);y--;
}
inc(i,l,mid)clr(a[i].y);
for(int x=l,y=mid+1;x<=mid||y<=r;)if(x<=mid&&(y>r||a[x].x<a[y].x)){
add(a[x].y);
tmp[cnt++]=a[x++];
}else{
a[y].v+=x-l-sum(a[y].y);
tmp[cnt++]=a[y++];
}
inc(i,l,mid)clr(a[i].y);
inc(i,l,r)a[i]=tmp[i];
}


int main(){
n=read();m=read();
inc(i,1,n)b[read()]=i;
dec(i,m,1){
a[++tot].id=i;a[tot].y=read();a[tot].x=b[a[tot].y];b[a[tot].y]=0;
}
inc(i,1,n)if(b[i])a[++tot]=P{0,b[i],i},b[i]=0;
reverse(a+1,a+1+tot);
cdq(1,tot);
sort(a+1,a+1+tot,cmp);
dec(i,tot,1)b[i]=b[i+1]+a[i].v;
inc(i,1,m)printf("%lld\n",b[i]);
return 0;
}