Problem 1049: Lost My Music【可持久化栈+倍增】

首先要求的式子是一个斜率的相反数，其实就是求斜率的最大值，那么我们只需维护一个下凸包即可，
考虑到直接用栈来存，如果是在一条链上的话可以保证每个数只会被插入弹出一次，直接做，暴力退栈就行了。
然而在树上暴力退的话会被卡成O（N2）……
所以对于每个点存一个倍增数组，记录其在凸壳里的祖先，然后乱搞即可……
代码：

//#include <bits/stdc++.h>#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <limits>#include <map>#include <vector>#include <queue> #define LL long long#define ft first#define sd second#define mp(x,y) make_pair(x,y)#define db double//#define int LLusing namespace std;const int N   = 5e5+10;//const int mod = ;const db INF =numeric_limits<double >::max();#define rep(i,x,y) for (int i=x;i<=y;++i)void read(int &x){    x=0;    char ch=getchar();    int f=1;    while (!isdigit(ch))       (ch=='-'?f=-1:0),ch=getchar();    while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();    x*=f;}int c[N],fa[N],dep[N];int head[N];struct xx{    int V,next;}b[N<<1];int G,n;void add(int x,int y){    b[++G]=(xx){y,head[x]};    head[x]=G;}db f[N];#define F(x) for (int i=head[x];i;i=b[i].next)#define v b[i].Vint q[N],hd,tl,nw;int anc[N][20],x;int ok(int xi,int yi,int xj,int yj,int xk,int yk){    return (LL)(yj-yi)*(LL)(xk-xj)<=(LL)(yk-yj)*(LL)(xj-xi);}signed main(){    freopen("data.txt","r",stdin);    freopen("Moon.txt","w",stdout);    read(n);    rep(i,1,n)         read(c[i]);    rep(i,2,n) read(fa[i]),add(fa[i],i);    //dfs(1,0);    q[1]=1;hd=tl=1;    while (hd<=tl)    {        nw=q[hd++];dep[nw]=dep[fa[nw]]+1;        F(nw) q[++tl]=v;        if (nw==1) continue;        x=fa[nw];        for (int i=2;i>=0;--i)        {            if (anc[ x][i]<=1) continue;            if (!ok(dep[ anc[anc[x][i]][0] ],c[ anc[anc[x][i]][0] ],dep[anc[x][i]],c[anc[x][i]],dep[nw],c[nw]))              x=anc[x][i];        }        if (x>1)          if (!ok(dep[anc[x][0]],c[anc[x][0]],dep[x],c[x],dep[nw],c[nw]))            x=anc[x][0];        anc[nw][0]=x;        rep(i,1,19) anc[nw][i]=anc[anc[nw][i-1]][i-1];    }    rep(i,2,n) printf("%.10f\n",(db)(c[anc[i][0]]-c[i])/(dep[i]-dep[anc[i][0]]));    return 0;}