【BZOJ1758】重建计划（点分治，长链剖分，线段树）

题意：

给定一棵n个点的树，每条边有权值。
求一条链，这条链包含的边数在L和U之间，且平均边权最大。
N﹤=100000

思路：

做法一：RYZ作业

二分答案再点分治，寻找是否有大于0且边数在L和U之间的链

f[i]为当前子树深度为i的链最大总和，g[i]为前几个深度为i的链最大总和

维护一个下标递增，值递增的单调队列

按子树深度排序一定要加，因为清空与当前子树深度最大值有关，不加的话可能会退化成n^2

改了快一天改不出，还加了迭代，最后发现是分治的时候root打成v，相当于什么都没干……

var head,vet,next,c,d,size,flag,q1:array[1..210000]of longint;
    p:array[0..210000]of longint;
    len,h,q2:array[1..210000]of double;
    f,g:array[0..210000]of double;
    n,i,tot,root,l1,r1,sum,mx,mxdep,x,y,z,tmp,t,w,now,m:longint;
    eps,oo,mid:double;

procedure swap(var x,y:longint);
var t:longint;
begin
 t:=x; x:=y; y:=t;
end;

function max(x,y:longint):longint;
begin
 if x>y then exit(x);
 exit(y);
end;

function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;

procedure add(a,b,c:longint);
begin
 inc(tot);
 next[tot]:=head[a];
 vet[tot]:=b;
 len[tot]:=c;
 head[a]:=tot;
end;

procedure getroot(u,fa:longint);
var e,v:longint;
begin
 size[u]:=1; p[u]:=0;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if (v<>fa)and(flag[v]=0) then
  begin
   getroot(v,u);
   size[u]:=size[u]+size[v];
   p[u]:=max(p[u],size[v]);
  end;
  e:=next[e];
 end;
 p[u]:=max(p[u],sum-p[u]);
 if p[u]<p[root] then root:=u;
end;

procedure dfs(u,fa,dep:longint;t:double);
var e,v:longint;
begin
 if f[dep]<t then f[dep]:=t;
 mx:=max(mx,dep);
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if (v<>fa)and(flag[v]=0) then dfs(v,u,dep+1,t+len[e]-mid);
  e:=next[e];
 end;

end;

procedure ins(k:longint;x:double);
begin
 while (w>=t)and(q2[w]<x) do dec(w);
 inc(w); q1[w]:=k; q2[w]:=x;
end;

procedure del(k:longint);
begin
 while (t<=w)and(q1[t]>=k) do inc(t);
end;

procedure getdep(u,fa,dep:longint);
var e,v:longint;
begin
 e:=head[u];
 if dep>d[m] then d[m]:=dep;
 while e<>0 do
 begin
  v:=vet[e];
  if (v<>fa)and(flag[v]=0) then getdep(v,u,dep+1);
  e:=next[e];
 end;
end;

procedure qsort(l,r:longint);
var i,j,mid:longint;
    t:double;
begin
 i:=l; j:=r; mid:=d[(l+r)>>1];
 repeat
  while mid>d[i] do inc(i);
  while mid<d[j] do dec(j);
  if i<=j then
  begin
   swap(d[i],d[j]);
   swap(c[i],c[j]);
   t:=h[i]; h[i]:=h[j]; h[j]:=t;
   inc(i); dec(j);
  end;
 until i>j;
 if l<j then qsort(l,j);
 if i<r then qsort(i,r);
end;

function solve(u:longint;avg:double):double;
var t1,t2,ans,tmp:double;
    v,e,i,j:longint;
begin
 t1:=avg; t2:=-oo;
 m:=0;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   inc(m);
   getdep(v,u,1);
   c[m]:=v; h[m]:=len[e];
  end;
  e:=next[e];
 end;
 if m>0 then qsort(1,m);
 while t1-t2>eps do
 begin
  t2:=t1; mid:=t1; ans:=-oo;
  mxdep:=0; g[0]:=0;
  for j:=1 to m do
  begin
   v:=c[j];
   mx:=0;
   dfs(v,u,1,h[j]-mid);
   t:=1; w:=0;
   for i:=min(mxdep,r1) to l1 do ins(i,g[i]);
   for i:=max(1,l1-mxdep) to mx do
   begin
    if l1-i>=0 then ins(l1-i,g[l1-i]);
    del(r1-i+1);
    if t<=w then
     if (f[i]+q2[t])/(q1[t]+i)>ans then
      ans:=(f[i]+q2[t])/(q1[t]+i);
   end;
   mxdep:=max(mxdep,mx);
   for i:=0 to mx do
    if f[i]>g[i] then g[i]:=f[i];
   for i:=0 to mx do f[i]:=-oo;
  end;
  for i:=0 to mxdep do g[i]:=-oo;
  t1:=t1+ans;
 end;
 flag[u]:=1; t2:=t1;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   root:=0; sum:=size[v];
   getroot(v,0);
   tmp:=solve(root,t1);
   if tmp>t2 then t2:=tmp;
  end;
  e:=next[e];
 end;
 exit(t2);

end;

begin
 assign(input,'bzoj1758.in'); reset(input);
 assign(output,'bzoj1758.out'); rewrite(output);
 readln(n);
 readln(l1,r1);
 oo:=1e8;
 for i:=1 to n-1 do
 begin
  readln(x,y,z);
  add(x,y,z);
  add(y,x,z);
 end;
 p[0]:=n+1; root:=0; sum:=n;
 eps:=1e-4;
 getroot(1,0);
 for i:=0 to n do
 begin
  f[i]:=-oo; g[i]:=-oo;
 end;
 writeln(solve(root,0):0:3);
 close(input);
 close(output);
end.

 做法2：From https://blog.bill.moe/WC2010-rebuild/

建立一个答案表，顺序是长链剖分的dfs序

长链可以共用答案表，轻儿子暴力合并到父亲所在长链中

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef pair<ll,int>P;
#define N  200010
#define M  200010
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=1e9+7,inv2=(MOD+1)/2;
      double eps=1e-4;
      int INF=1<<30;
      ll inf=5e13;
      int dx[4]={-1,1,0,0};
      int dy[4]={0,0,-1,1};

double t[N<<2];
int head[N],vet[N],len[N],nxt[N],
    dep[N],son[N],slen[N],fa[N],id[N],dfn[N],mx[N],top[N],
    tot,L,R,tim,n;
double ans,tmp[N];

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

void add(int a,int b,int c)
{
    nxt[++tot]=head[a];
    vet[tot]=b;
    len[tot]=c;
    head[a]=tot;
}

void build(int l,int r,int p)
{
    t[p]=-1e18;
    if(l==r)
    {
        id[l]=p;
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,ls);
    build(mid+1,r,rs);
}

void update(int u,double v)
{
    int now=id[u];
    while(now)
    {
        t[now]=max(t[now],v);
        now>>=1;
    }
}

double query(int l,int r,int x,int y,int p)
{
    if(x<=l&&r<=y) return t[p];
    int mid=(l+r)>>1;
    double res=-1e18;
    if(x<=mid) res=max(res,query(l,mid,x,y,ls));
    if(y>mid) res=max(res,query(mid+1,r,x,y,rs));
    return res;
}

void dfs1(int u,int pre,int d)
{
    dep[u]=mx[u]=d;
    fa[u]=pre;
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=pre)
        {
            dfs1(v,u,d+1);
            if(mx[v]>mx[son[u]])
            {
                son[u]=v;
                slen[u]=len[e];
                mx[u]=mx[v];
            }
        }
        e=nxt[e];
    }
}

void dfs2(int u,int ance)
{
    top[u]=ance;
    dfn[u]=++tim;
    if(son[u]) dfs2(son[u],ance);
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=fa[u]&&v!=son[u]) dfs2(v,v);
        e=nxt[e];
    }
}

double ask(int u,int l,int r)
{
    l=max(l,0);
    r=min(r,mx[u]-dep[u]);
    if(l>r) return -1e18;
    return query(1,n,dfn[u]+l,dfn[u]+r,1);
}

void solve(int u,double d,double mid)
{
    update(dfn[u],d);
    if(son[u]) solve(son[u],d+slen[u]-mid,mid);
    int e=head[u];
    while(e)
    {
        int v=vet[e];
        if(v!=fa[u]&&v!=son[u])
        {
            solve(v,d+len[e]-mid,mid);
            rep(j,1,mx[v]-dep[v]+1)
            {
                tmp[j]=t[id[dfn[v]+j-1]];
                ans=max(ans,tmp[j]+ask(u,L-j,R-j)-2*d);
            }
            rep(j,1,mx[v]-dep[v]+1) update(dfn[u]+j,tmp[j]);
        }
        e=nxt[e];
    }
    ans=max(ans,ask(u,L,R)-d);
}

int isok(double K)
{
    build(1,n,1);
    ans=-1e18;
    solve(1,0,K);
    return ans>=-eps;
}

int main()
{
    n=read(),L=read(),R=read();
    tot=0;
    rep(i,1,n-1)
    {
        int x=read(),y=read(),z=read();
        add(x,y,z);
        add(y,x,z);
    }
    dfs1(1,0,1);
    tim=0;
    dfs2(1,1);
    double left=0,right=1e10;
    while(right-left>eps)
    {
        double mid=(left+right)/2;
        if(isok(mid)) left=mid;
         else right=mid;
    }
    printf("%.3f\n",left);
    return 0;
}