题目大意
\(n\)(\(n\leq10^5\))个点的一棵树，有边权\(w\)，给定\(l,r\)，求边数在\([l,r]\)中的路径的平均边权的最大值
题解
二分答案，判断时将边权变成\(w-mid\)，判断是否存在一条边权总和非负且边数在[l,r]的路径
设\(f(i,j)\)表示从点\(i\)往下走\(j\)条边的边权总和最多是多少
则有\(f(i,j)=max_{v\in son(i)}\{f(v,j-1)+w(i,v)\}\)
\(ans(i)=max_{v\in son(i) , l\leq k+j+1\leq r}\{f(v,k)+f(i,j)+w(i,v)\}\)
考虑长链剖分，将\(f(i,j)\)存到\(F(i,j)=dfn[i]+j\)的位置
需要注意的是，算完点\(i\)的长儿子后，需要将\(F(i,1)\)到\(F(i,dep[i])\)的值+\(w(i,v)\)
看上去常数比点分治小，事实上。。。
代码
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define view(u,k) for(int k=fir[u];~k;k=nxt[k])
#define maxn 100010
#define maxm (maxn<<1)
#define F(x,y) (dfn[x]+y)
#define inf (1e11)
#define ls (u<<1)
#define rs (u<<1|1)
#define mi (L+R>>1)
#define eps (1e-4)
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)&&ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
void write(int x)
{
if(x==0){putchar('0'),putchar('\n');return;}
int f=0;char ch[20];
if(x<0)putchar('-'),x=-x;
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
putchar('\n');
return;
}
int n,l,r,cnt,fir[maxn],nxt[maxm],v[maxm],son[maxn],fa[maxn],dfn[maxn],tim,dep[maxn];
double toso[maxn],w[maxm],tr[maxn<<2],mk[maxn<<2],ans,fakeans,mid;
inline void ade(int u1,int v1,int w1){w[cnt]=(double)w1,v[cnt]=v1,nxt[cnt]=fir[u1],fir[u1]=cnt++;}
void getdep(int u)
{
view(u,k)if(v[k]!=fa[u])
{
fa[v[k]]=u,getdep(v[k]),dep[u]=max(dep[v[k]]+1,dep[u]);
if(!son[u]||dep[v[k]]>dep[son[u]])son[u]=v[k],toso[u]=w[k];
}
}
void getdfn(int u)
{
dfn[u]=++tim;
if(son[u])getdfn(son[u]);
view(u,k)if(v[k]!=fa[u]&&v[k]!=son[u])getdfn(v[k]);
}
inline void mark(int u,double k){if(tr[u]!=-inf)mk[u]+=k,tr[u]+=k;}
inline void pd(int u){if(mk[u]==-inf){tr[ls]=tr[rs]=-inf,mk[ls]=mk[rs]=-inf,mk[u]=0;}if(mk[u]){mark(ls,mk[u]),mark(rs,mk[u]),mk[u]=0;}}
void add(int u,int L,int R,int x,int y,double k)
{
if(y<L||R<x||y<x)return ;
if(x<=L&&R<=y)return mark(u,k);
pd(u);
if(x<=mi)add(ls,L,mi,x,y,k);
if(y>mi)add(rs,mi+1,R,x,y,k);
tr[u]=max(tr[ls],tr[rs]);
}
void chg(int u,int L,int R,int x,double k)
{
if(x<L||R<x)return;
if(x<=L&&R<=x){tr[u]=max(tr[u],k);return;}
pd(u);
if(x<=mi)chg(ls,L,mi,x,k);
if(x>mi)chg(rs,mi+1,R,x,k);
tr[u]=max(tr[ls],tr[rs]);
}
double ask(int u,int L,int R,int x,int y)
{
if(y<L||R<x||y<x)return -inf;
if(x<=L&&R<=y)return tr[u];
pd(u);
double res=-inf;
if(x<=mi)res=ask(ls,L,mi,x,y);
if(y>mi)res=max(res,ask(rs,mi+1,R,x,y));
return res;
}
double tmp[maxn];
void getans(int u)
{
if(son[u])getans(son[u]),add(1,1,n,F(u,1),F(u,dep[u]),toso[u]-mid);
if(fakeans>=0)return;
chg(1,1,n,F(u,0),0);
int Llim=max(0,l),Rlim=min(dep[u],r);
if(Llim<=Rlim)
{
double tmp=ask(1,1,n,F(u,Llim),F(u,Rlim));
fakeans=max(fakeans,tmp);
if(fakeans>=0)return;
}
view(u,k)if(v[k]!=fa[u]&&v[k]!=son[u])
{
getans(v[k]);
if(fakeans>=0)return;
rep(i,0,dep[v[k]])
{
Llim=max(0,l-i-1),Rlim=min(r-i-1,dep[u]);
tmp[i]=ask(1,1,n,F(v[k],i),F(v[k],i))+w[k]-mid;
if(Llim<=Rlim)
{
double tmp2=ask(1,1,n,F(u,Llim),F(u,Rlim));
fakeans=max(fakeans,tmp2+tmp[i]);
if(fakeans>=0)return;
}
}
rep(i,0,dep[v[k]])chg(1,1,n,F(u,i+1),tmp[i]);
}
}
int main()
{
memset(fir,-1,sizeof(fir));
n=read(),l=read(),r=read();double L=inf,R=0;
rep(i,1,n-1){int x=read(),y=read(),z=read();ade(x,y,z),ade(y,x,z),L=min(L,(double)z),R=max(R,(double)z);}
getdep(1),getdfn(1);
while(R-L>=eps)
{
mid=(R+L)/2.0;
fakeans=-inf;tr[1]=mk[1]=-inf;
getans(1);
if(fakeans>=0)ans=max(ans,mid),L=mid+eps;
else R=mid-eps;
}
printf("%.3lf",ans);
return 0;
}
/*
4
2 3
1 2 1
1 3 2
1 4 3
*/