【HDOJ4812】D Tree（点分治）

题意：

给定一棵 n 个点的树,每个点有权值 Vi

问是否存在一条路径使得路径上所有点的权值乘积 mod(10^6 + 3) 为 K

输出路径的首尾标号,若有多解,输出字典序最小的解

对于100%的数据,有1≤n≤10^5,0≤K≤10^6+2,1≤vi ≤10^6+2

思路：RYZ作业

预处理逆元

哈希路径权值乘积的模数，保存编号用来查询

 const mo=;
 var head,vet,next,len,flag,son:array[..]of longint;
     f:array[..]of longint;
     dis,a:array[..]of int64;
     q:array[..,..]of longint;
     exf:array[..mo]of int64;
     hash:array[..mo]of longint;
     n,m,i,j,tot,x,y,z,ans1,ans2,sum,root,top,k:longint;
 
 procedure add(a,b:longint);
 begin
  inc(tot);
  next[tot]:=head[a];
  vet[tot]:=b;
  head[a]:=tot;
 end;
 
 function max(x,y:longint):longint;
 begin
  if x>y then exit(x);
  exit(y);
 end;
 
 procedure swap(var x,y:longint);
 var t:longint;
 begin
  t:=x; x:=y; y:=t;
 end;
 
 function getroot(u,fa:longint):longint;
 var e,v:longint;
 begin
  son[u]:=; f[u]:=;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if (v<>fa)and(flag[v]=) then
   begin
    getroot(v,u);
    son[u]:=son[u]+son[v];
    f[u]:=max(f[u],son[v]);
   end;
   e:=next[e];
  end;
  f[u]:=max(f[u],sum-f[u]);
  if f[u]<f[root] then root:=u;
 end;
 
 procedure dfs(u,fa:longint);
 var e,v:longint;
 begin
  inc(top); q[top,]:=dis[u]; q[top,]:=u;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if (v<>fa)and(flag[v]=) then
   begin
    dis[v]:=dis[u]*a[v] mod mo;
    dfs(v,u);
   end;
   e:=next[e];
  end;
 end;
 
 procedure cmp(x,id:longint);
 var y:longint;
 begin
  x:=exf[x]*k mod mo;
  y:=hash[x];
  if y= then exit;
  if y>id then swap(y,id);
  if (y<ans1)or((y=ans1)and(id<ans2)) then
  begin
   ans1:=y; ans2:=id;
  end;
 end;
 
 procedure solve(u:longint);
 var e,v,i,now:longint;
 begin
  flag[u]:=; hash[a[u]]:=u;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    top:=; dis[v]:=a[v];
    dfs(v,u);
    for i:= to top do cmp(q[i,],q[i,]); //对于每一个子树，因为不能重复计算u这个根结点，分别计算子树内结果，后面再用加上根结点的答案更新
    top:=; dis[v]:=a[u]*a[v] mod mo;
    dfs(v,u);
    for i:= to top do
    begin
     now:=hash[q[i,]];
     if (now=)or(q[i,]<now) then hash[q[i,]]:=q[i,]; //用子树中的路径加上根结点，更新经过根结点的答案
    end;
   end;
   e:=next[e];
  end;
  hash[a[u]]:=;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    top:=; dis[v]:=a[u]*a[v] mod mo;
    dfs(v,u);
    for i:= to top do hash[q[i,]]:=; //清空所有信息
   end;
   e:=next[e];
  end;
  e:=head[u];
  while e<> do
  begin
   v:=vet[e];
   if flag[v]= then
   begin
    root:=; sum:=son[v];
    getroot(v,);
    solve(root);
   end;
   e:=next[e];
  end;
 end;
 
 begin
  assign(input,'hdoj4812.in'); reset(input);
  assign(output,'hdoj4812.out'); rewrite(output);
  exf[]:=; exf[]:=;
  for i:= to  do exf[i]:=exf[mo mod i]*(mo-mo div i) mod mo;
  while not eof do
  begin
  // fillchar(head,sizeof(head),);
   //fillchar(flag,sizeof(flag),);
   for i:= to n do
   begin
    head[i]:=; flag[i]:=;
   end;
   tot:=; ans1:=maxlongint; ans2:=maxlongint;
   read(n,k);
   if n= then break;
   for i:= to n do read(a[i]);
   for i:= to n- do
   begin
    read(x,y);
    add(x,y);
    add(y,x);
   end;
   root:=; sum:=n; f[]:=n+;
   getroot(,);
   solve(root);
   if ans1=maxlongint then writeln('No solution')
    else writeln(ans1,' ',ans2);
  end;
  close(input);
  close(output);
 end.