这算是分数规划的一种...?打扰了...
先二分答案，问题转为判定是否存在$\frac{y_i+q_j}{x_i+p_j}\geq k$
整理得$-x_ik+y_i-p_jk+q_j\geq0$，于是我们现在要分别找到（$(x_i,y_i)$满足$-x_ik+y_i$最大）和（$(p_j,q_j)$满足$-p_jk+q_j$最大）并判断是否满足此式
把这些参数看成直线，直接树剖+线段树维护直线形成的下凸壳即可，查询就在凸壳上二分，总时间复杂度$O(n\log_2^3n\log_2\text{ans})$（二分答案，重链，线段树，凸包上二分），因为后三个$\log$都很难跑满，所以能过
#include<stdio.h>
#include<algorithm>
using namespace std;
typedef double du;
const du eps=1e-4,inf=2147483647;
bool leq(du a,du b){return a-b<=eps;}
struct line{
du k,b;
line(du x=0,du y=0){k=x;b=y;}
du val(du x){return k*x+b;}
}l[2][30010],t[30010];
typedef line*linep;
linep ch[2][120010];
bool operator<(line a,line b){return a.k==b.k?a.b>b.b:a.k<b.k;}
du its(line a,line b){
return(b.b-a.b)/(a.k-b.k);
}
int h[30010],to[60010],nex[60010],M,n;
void add(int a,int b){
M++;
to[M]=b;
nex[M]=h[a];
h[a]=M;
}
int fa[30010],dep[30010],siz[30010],son[30010],pos[30010],bl[30010],rpos[30010];
void dfs(int x){
int i,k=0,mx=0;
dep[x]=dep[fa[x]]+1;
siz[x]=1;
for(i=h[x];i;i=nex[i]){
if(to[i]!=fa[x]){
fa[to[i]]=x;
dfs(to[i]);
siz[x]+=siz[to[i]];
if(siz[to[i]]>mx){
mx=siz[to[i]];
k=to[i];
}
}
}
son[x]=k;
}
void dfs(int x,int chain){
bl[x]=chain;
pos[x]=++M;
rpos[M]=x;
if(son[x])dfs(son[x],chain);
for(int i=h[x];i;i=nex[i]){
if(to[i]!=fa[x]&&to[i]!=son[x])dfs(to[i],to[i]);
}
}
int build(linep&h,linep a,int l,int r){
int n,i,top;
n=r-l+1;
for(i=l;i<=r;i++)t[i-l+1]=a[rpos[i]];
sort(t+1,t+n+1);
top=0;
for(i=1;i<=n;i++){
if(t[i].k==t[i-1].k)continue;
while(top>1&&leq(its(t[top-1],t[i]),its(t[top-1],t[top])))top--;
top++;
t[top]=t[i];
}
h=new line[top+1];
for(i=1;i<=top;i++)h[i]=t[i];
return top;
}
int len[2][120010];
void build(int l,int r,int x){
len[0][x]=build(ch[0][x],::l[0],l,r);
len[1][x]=build(ch[1][x],::l[1],l,r);
if(l==r)return;
int mid=(l+r)>>1;
build(l,mid,x<<1);
build(mid+1,r,x<<1|1);
}
du query(du k,int n,line*ch){
int l,r,mid,i;
l=1;
r=n;
while(r-l>3){
mid=(l+r)>>1;
if(leq(k,its(ch[mid],ch[mid-1])))
r=mid;
else
l=mid;
}
du res=-inf;
for(i=l;i<=r;i++)res=max(res,ch[i].val(k));
return res;
}
du query(int f,int L,int R,du k,int l,int r,int x){
if(L<=l&&r<=R)return query(k,len[f][x],ch[f][x]);
int mid=(l+r)>>1;
du ans=-inf;
if(L<=mid)ans=max(ans,query(f,L,R,k,l,mid,x<<1));
if(mid<R)ans=max(ans,query(f,L,R,k,mid+1,r,x<<1|1));
return ans;
}
du calc(int f,int x,int y,du k){
du ans=-inf;
while(bl[x]!=bl[y]){
if(dep[bl[x]]<dep[bl[y]])swap(x,y);
ans=max(ans,query(f,pos[bl[x]],pos[x],k,1,n,1));
x=fa[bl[x]];
}
if(pos[x]>pos[y])swap(x,y);
return max(ans,query(f,pos[x],pos[y],k,1,n,1));
}
du solve(int x,int y){
du l,r,mid,ans,t;
l=0;
r=100000;
while(r-l>eps){
mid=(l+r)*.5;
t=calc(0,x,y,mid)+calc(1,x,y,mid);
if(leq(0,t)){
l=mid;
ans=mid;
}else
r=mid;
}
return ans;
}
int main(){
int m,i,x,y;
du t;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%lf",&t);
l[0][i].k=-t;
}
for(i=1;i<=n;i++)scanf("%lf",&l[0][i].b);
for(i=1;i<=n;i++){
scanf("%lf",&t);
l[1][i].k=-t;
}
for(i=1;i<=n;i++)scanf("%lf",&l[1][i].b);
for(i=1;i<n;i++){
scanf("%d%d",&x,&y);
add(x,y);
add(y,x);
}
dfs(1);
M=0;
dfs(1,1);
build(1,n,1);
scanf("%d",&m);
while(m--){
scanf("%d%d",&x,&y);
printf("%.4lf\n",solve(x,y));
}
}