【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=1000003;
var head,vet,next,len,flag,son:array[1..210000]of longint;
    f:array[0..110000]of longint;
    dis,a:array[0..110000]of int64;
    q:array[1..110000,1..2]of longint;
    exf:array[0..mo]of int64;
    hash:array[0..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]:=1; f[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);
   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,1]:=dis[u]; q[top,2]:=u;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if (v<>fa)and(flag[v]=0) 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=0 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]:=1; hash[a[u]]:=u;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   top:=0; dis[v]:=a[v];
   dfs(v,u);
   for i:=1 to top do cmp(q[i,1],q[i,2]); //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:=0; dis[v]:=a[u]*a[v] mod mo;
   dfs(v,u);
   for i:=1 to top do
   begin
    now:=hash[q[i,1]];
    if (now=0)or(q[i,2]<now) then hash[q[i,1]]:=q[i,2]; //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]]:=0;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   top:=0; dis[v]:=a[u]*a[v] mod mo;
   dfs(v,u);
   for i:=1 to top do hash[q[i,1]]:=0; //清空所有信息
  end;
  e:=next[e];
 end;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if flag[v]=0 then
  begin
   root:=0; sum:=son[v];
   getroot(v,0);
   solve(root);
  end;
  e:=next[e];
 end;
end;

begin
 assign(input,'hdoj4812.in'); reset(input);
 assign(output,'hdoj4812.out'); rewrite(output);
 exf[0]:=1; exf[1]:=1;
 for i:=2 to 1000003 do exf[i]:=exf[mo mod i]*(mo-mo div i) mod mo;
 while not eof do
 begin
 // fillchar(head,sizeof(head),0);
  //fillchar(flag,sizeof(flag),0);
  for i:=1 to n do
  begin
   head[i]:=0; flag[i]:=0;
  end;
  tot:=0; ans1:=maxlongint; ans2:=maxlongint;
  read(n,k);
  if n=0 then break;
  for i:=1 to n do read(a[i]);
  for i:=1 to n-1 do
  begin
   read(x,y);
   add(x,y);
   add(y,x);
  end;
  root:=0; sum:=n; f[0]:=n+1;
  getroot(1,0);
  solve(root);
  if ans1=maxlongint then writeln('No solution')
   else writeln(ans1,' ',ans2);
 end;
 close(input);
 close(output);
end.