注意到用这个集合产生方式可以构建出一个树型结构,并且每个加入/删除元素都是对应的一个子树的范围,对应到dfs序上就是每次对一个区间内的集合加入/删除元素,所以可以线段树分治,把每种元素的出现区间整出来
把答案柿子((x-x_0)^2+c)拆开,得到(x^2-2x*x_0+{x_0}^2+c),然后每个元素就相当于斜率为(-2x),截距为(x^2+c)的直线,所以线段树每个节点维护凸壳,要用的时候直接查对应横坐标的值.如果直接做是两个(log)的,但是因为要答案最小,所以可以按照斜率从大到小((x)从小到大)的顺序插入元素对应的直线,然后按照(x_0)从小到大查询,这样由于插入和查询的单调性所以可以少掉一个log
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define LL long long
#define uLL unsigned long long
#define db double
using namespace std;
const int N=5e5+10;
LL rd()
{
LL x=0,w=1;char ch=0;
while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
return x*w;
}
struct node
{
LL x,y;
}a[N],qr[N];
vector<int> ti[N],e[N];
LL an[N];
int n,m,q,fa[N],ty[N],sz[N],dfn[N],tt,sb[N];
bool cmp1(int aa,int bb){return a[aa].x!=a[bb].x?a[aa].x<a[bb].x:a[aa].x*a[aa].x+a[aa].y>a[bb].x*a[bb].x+a[bb].y;}
bool cmp2(int aa,int bb){return qr[aa].x<qr[bb].x;}
void dfs(int x)
{
dfn[x]=++tt,sz[x]=1;
ti[ty[x]].push_back(dfn[x]);
vector<int>::iterator it;
for(it=e[x].begin();it!=e[x].end();++it)
{
int y=*it;
dfs(y),sz[x]+=sz[y];
}
ti[ty[x]].push_back(dfn[x]+sz[x]);
}
struct line
{
db k,b;
line(){}
line(LL x,LL y){k=-2*x,b=x*x+y;}
};
db jiao(line aa,line bb){return (bb.b-aa.b)/(aa.k-bb.k);}
struct HULL
{
vector<line> qu;
int hd,tl;
HULL(){hd=0,tl=-1;}
void inst(line aa)
{
if(hd<=tl&&qu[tl].k==aa.k) qu.pop_back(),--tl;
while(hd<tl&&jiao(aa,qu[tl])<=jiao(qu[tl],qu[tl-1])) qu.pop_back(),--tl;
qu.push_back(aa),++tl;
}
LL quer(LL x)
{
while(hd<tl&&x>=jiao(qu[hd],qu[hd+1])) ++hd;
return hd<=tl?(LL)round(qu[hd].k*(db)x+qu[hd].b):1ll<<50;
}
}hl[N<<2];
#define mid ((l+r)>>1)
int ps[N];
void setli(int o,int l,int r,int ll,int rr,line x)
{
if(ll<=l&&r<=rr){hl[o].inst(x);return;}
if(ll<=mid) setli(o<<1,l,mid,ll,rr,x);
if(rr>mid) setli(o<<1|1,mid+1,r,ll,rr,x);
}
void bui(int o,int l,int r)
{
if(l==r){ps[l]=o;return;}
bui(o<<1,l,mid),bui(o<<1|1,mid+1,r);
}
int main()
{
n=rd(),q=rd();
a[++m]=(node){0,rd()};
ty[1]=1;
for(int i=2;i<=n;++i)
{
int op=rd();
fa[i]=rd()+1;
int ii=rd()+1;
e[fa[i]].push_back(i);
ty[i]=ii;
m=max(m,ii);
if(op==0)
{
a[ii].x=rd();
rd(),rd();
a[ii].y=rd();
}
}
dfs(1);
for(int i=1;i<=n;++i) sb[i]=i;
sort(sb+1,sb+n+1,cmp1);
for(int i=1;i<=n;++i)
{
int ii=sb[i],nn=ti[ii].size();
for(int j=0;j+1<nn;j+=2)
if(ti[ii][j]<=ti[ii][j+1]-1)
setli(1,1,n,ti[ii][j],ti[ii][j+1]-1,line(a[ii].x,a[ii].y));
}
for(int i=1;i<=q;++i)
{
int y=rd()+1,x=rd();
qr[i]=(node){x,y};
}
for(int i=1;i<=q;++i) sb[i]=i;
sort(sb+1,sb+q+1,cmp2);
bui(1,1,n);
for(int i=1;i<=q;++i)
{
int ii=sb[i],es=qr[ii].y;
LL y=qr[ii].x;
int o=ps[dfn[es]];
an[ii]=1ll<<50;
while(o)
{
an[ii]=min(an[ii],y*y+hl[o].quer(y));
o>>=1;
}
}
for(int i=1;i<=q;++i) printf("%lld
",an[i]);
return 0;
}