【2020省選模擬】時間複雜度（啓發式分裂）

$3.14$

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由於$b$不單增
考慮一個做法
每次把區間所有出現次數小於$b_{r-l+1}$的刪去
然後遞歸子區間

發現沒必要刪所有
每次找一個刪去遞歸就可以
可以維護着每個出現次數
左右一起啓發式分類

複雜度就是$O(nlogn)$

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pb push_back
#define pii pair<int,int>
#define ll long long
#define fi first
#define se second
#define bg begin
cs int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ib==ob)?EOF:*ib++;
}
inline int read(){
    char ch=gc();
    int res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline ll readll(){
    char ch=gc();
    ll res=0;bool f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
inline int readstring(char *s){
	int top=0;char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<typename tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<typename tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
typedef vector<int> poly;
cs int N=1000005;
int a[N],n,b[N],vis[N],ans;
void solve(int l,int r){
	if(l>r)return;
	int len=r-l+1,pos=0,fg=0;
	for(int ls=l,rs=r;ls<=rs;ls++,rs--){
		if(vis[a[ls]]<b[len]){fg=0;pos=ls;break;}
		if(vis[a[rs]]<b[len]){fg=1;pos=rs;break;}
	}
	if(!pos){chemx(ans,len);for(int i=l;i<=r;i++)vis[a[i]]--;return;}
	if(fg==0){
		for(int i=l;i<=pos;i++)vis[a[i]]--;
		solve(pos+1,r);
		for(int i=l;i<pos;i++)vis[a[i]]++;
		solve(l,pos-1);
	}
	else {
		for(int i=pos;i<=r;i++)vis[a[i]]--;
		solve(l,pos-1);
		for(int i=pos+1;i<=r;i++)vis[a[i]]++;
		solve(pos+1,r);
	}
}
int main(){
	#ifdef Stargazer
	freopen("lx.in","r",stdin);
	#endif
	n=read();
	for(int i=1;i<=n;i++)a[i]=read(),vis[a[i]]++;
	for(int i=1;i<=n;i++)b[i]=read();
	solve(1,n);
	cout<<ans<<'\n';
	return 0;
}