【BZOJ1758】重建计划（点分治）

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[..]of longint;
     p:array[..]of longint;
     len,h,q2:array[..]of double;
     f,g:array[..]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]:=; p[u]:=;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if (v<>fa)and(flag[v]=) 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<> do
  begin
   v:=vet[e];
   if (v<>fa)and(flag[v]=) then dfs(v,u,dep+,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<> do
  begin
   v:=vet[e];
   if (v<>fa)and(flag[v]=) then getdep(v,u,dep+);
   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)>>];
  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:=;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    inc(m);
    getdep(v,u,);
    c[m]:=v; h[m]:=len[e];
   end;
   e:=next[e];
  end;
  if m> then qsort(,m);
  while t1-t2>eps do
  begin
   t2:=t1; mid:=t1; ans:=-oo;
   mxdep:=; g[]:=;
   for j:= to m do
   begin
    v:=c[j];
    mx:=;
    dfs(v,u,,h[j]-mid);
    t:=; w:=;
    for i:=min(mxdep,r1) to l1 do ins(i,g[i]);
    for i:=max(,l1-mxdep) to mx do
    begin
     if l1-i>= then ins(l1-i,g[l1-i]);
     del(r1-i+);
     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:= to mx do
     if f[i]>g[i] then g[i]:=f[i];
    for i:= to mx do f[i]:=-oo;
   end;
   for i:= to mxdep do g[i]:=-oo;
   t1:=t1+ans;
  end;
  flag[u]:=; t2:=t1;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    root:=; sum:=size[v];
    getroot(v,);
    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:= to n- do
  begin
   readln(x,y,z);
   add(x,y,z);
   add(y,x,z);
  end;
  p[]:=n+; root:=; sum:=n;
  eps:=1e-4;
  getroot(,);
  for i:= to n do
  begin
   f[i]:=-oo; g[i]:=-oo;
  end;
  writeln(solve(root,)::);
  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+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=<<;
       ll inf=5e13;
       int dx[]={-,,,};
       int dy[]={,,-,};

 double t[N<<];
 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=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,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)>>;
     build(l,mid,ls);
     build(mid+,r,rs);
 }

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

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

 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,,mx[v]-dep[v]+)
             {
                 tmp[j]=t[id[dfn[v]+j-]];
                 ans=max(ans,tmp[j]+ask(u,L-j,R-j)-*d);
             }
             rep(j,,mx[v]-dep[v]+) update(dfn[u]+j,tmp[j]);
         }
         e=nxt[e];
     }
     ans=max(ans,ask(u,L,R)-d);
 }

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

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