題解 Lost My Music

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• 可持久化單調棧/隊列注意要倍增處理

  這裡涉及求凸包切線

  如果是個普通單調棧維護的凸包直接三分求就行

  但這裡的可持久化單調棧是倍增維護的，不好三分

  是以考慮如何判斷目前枚舉到的點在最終答案點的哪一側

  如果我們要check一個點v，發現在答案左側有calc(u, fa(v))>calc(u, v)，而答案右側有calc(u, fa(v))<calc(u, v)

  就可以二分了 其實普通單調棧也可以這樣避免三分

  完全想不到……

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000100
#define ll long long 
#define ld long double
#define usd unsigned
#define ull unsigned long long
#define double long double 
//#define int long long 

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int head[N], size;
double c[N];
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {edge* k=&e[++size]; k->to=t; k->next=head[s]; head[s]=size;}

namespace force{
	double ans[N];
	void dfs(int u, double cv, int dis) {
		if (dis) ans[u] = min(ans[u], (cv-c[u])/(1.0*dis));
		for (int i=head[u]; i; i=e[i].next) dfs(e[i].to, cv, dis+1);
	}
	void solve() {
		for (int i=1; i<=n; ++i) ans[i]=1e30;
		for (int i=1; i<=n; ++i) dfs(i, c[i], 0);
		for (int i=2; i<=n; ++i) printf("%.10Lf\n", ans[i]);
		exit(0);
	}
}

namespace task{
	//struct que{double c, d, k; inline void build(double c_, double d_, double k_){c=c_; d=d_; k=k_;}}sta[N][25];
	int sta[N][25];
	double ans[N], dep[N], reck[N];
	inline double calc(int u, int v) {return (c[u]-c[v])/(dep[u]-dep[v]);}
	void dfs(int u) {
		//cout<<"dfs "<<u<<endl;
		//cout<<"P: "<<dep[u]<<' '<<c[u]<<endl;
		for (int i=1; i<=22; ++i) 
			if (dep[u]>=(1<<i)) sta[u][i]=sta[sta[u][i-1]][i-1];
			else break;
		//cout<<"sta: "; for (int t=sta[u][0]; t; t=sta[t][0]) cout<<t<<' '; cout<<endl;
		if (u!=1) {
			int now=u;
			for (int i=22; i>=0; --i) 
				if (sta[sta[now][i]][0] && calc(u, sta[now][i])<calc(u, sta[sta[now][i]][0])) now=sta[now][i]; // cout<<"now: "<<now<<endl;
			//while (sta[now][0] && calc(u, sta[now][0])>=k) {k=calc(u, sta[now][0]); now=sta[now][0];} //cout<<"now: "<<now<<' '<<k<<endl;
			if (sta[now][0]) now=sta[now][0];
			ans[u]=calc(u, now);
			//cout<<"choose: "<<now<<endl;
			sta[u][0]=now;
		}
		for (int i=1; i<=22; ++i) 
			if (dep[u]>=(1<<i)) sta[u][i]=sta[sta[u][i-1]][i-1];
			else break;
		for (int i=head[u],v; i; i=e[i].next) {
			v = e[i].to;
			dep[v]=dep[u]+1; sta[v][0]=u;
			dfs(v);
		}
		//cout<<"return "<<endl;
	}
	void solve() {
		dep[1]=1;
		dfs(1);
		for (int i=2; i<=n; ++i) printf("%.10Lf\n", -1.0*ans[i]);
		exit(0);
	}
}

signed main()
{
	#ifdef DEBUG
	freopen("1.in", "r", stdin);
	#endif
	
	n=read();
	for (int i=1; i<=n; ++i) c[i]=read();
	for (int i=2; i<=n; ++i) add(read(), i);
	//force::solve();
	task::solve();

	return 0;
}