Description
Solution
题目就是求直径,
先排个欧拉序,这样每棵子树的编号范围就可知了,
用线段树预处理出每个编号区间的直径,
每次的询问,设LCA(x,y)=q,a.en为以a为根的子树中的最大编号,
查询就是(1~a的编号-1),(a.en+1~b的编号-1),(b.en+1~n)这几个区间的直径,再合并一下即可
复杂度:O(nlog2(n))
Code
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define efo(i,q) for(int i=A[q];i;i=B[i][0])
#define iff() if(B[i][1]!=fa)
#define rmx(q,w,j) (a[rmq[q][j]].c>a[rmq[w][j]].c?rmq[w][j]:rmq[q][j])
#define MRG(q,w) if((t=LCAs(q,w))>s.s)s.a=q,s.b=w,s.s=t
using namespace std;
const int N=100500;
int read(int &n)
{
char ch=' ';int q=0,w=1;
for(;(ch!='-')&&((ch<'0')||(ch>'9'));ch=getchar());
if(ch=='-')w=-1,ch=getchar();
for(;ch>='0' && ch<='9';ch=getchar())q=q*10+ch-48;n=q*w;return n;
}
int n,m,ans;
int B[2*N][2],A[N],B0=1;
struct qqww
{int c,zx,ad,en;}a[N];
int zx[N];
int rmq[N*2][20],rmq0,er[20];
struct wwqq
{int a,b,s;}b[N*4];
int max(int a,int b){return a>b?a:b;}
void link(int q,int w)
{
B[++B0][0]=A[q],A[q]=B0,B[B0][1]=w;
B[++B0][0]=A[w],A[w]=B0,B[B0][1]=q;
}
int LCA(int q,int w)
{
q=a[q].ad,w=a[w].ad;if(q>w)swap(q,w);
int t=1.0*log(w-q+1)/log(2);
return rmx(q,w-er[t]+1,t);
}
int LCAs(int q,int w){return a[q].c+a[w].c-2*a[LCA(q,w)].c;}
int dfs1(int q,int fa,int c)
{
zx[a[q].zx=++zx[0]]=q;a[q].c=c;
rmq[a[q].ad=++rmq0][0]=q;
efo(i,q)iff()dfs1(B[i][1],q,c+1),rmq[++rmq0][0]=q;
a[q].en=zx[0];
}
wwqq merge(wwqq Q,wwqq W)
{
if(!Q.a)return W;if(!W.a)return Q;
wwqq s;int t;
int q=Q.a,w=Q.b,e=W.a,r=W.b;
s.a=q,s.b=w,s.s=Q.s;
if(W.s>s.s)s.a=e,s.b=r,s.s=W.s;
MRG(q,e);MRG(q,r);MRG(w,e);MRG(w,r);
return s;
}
wwqq merge(wwqq q,wwqq w,wwqq e){return merge(q,merge(w,e));}
void build(int l,int r,int e)
{
if(l==r){b[e].a=b[e].b=zx[l],b[e].s=0;return;}
int t=(l+r)/2;
build(l,t,e*2),build(t+1,r,e*2+1);
b[e]=merge(b[e*2],b[e*2+1]);
}
wwqq find(int l,int r,int e,int l1,int r1)
{
if(!l1)l1=1;if(l1>r1)return b[0];
if(l==l1&&r==r1)return b[e];
int t=(l+r)/2;
if(r1<=t)return find(l,t,e*2,l1,r1);
else if(t<l1)return find(t+1,r,e*2+1,l1,r1);
else return merge(find(l,t,e*2,l1,t),find(t+1,r,e*2+1,t+1,r1));
}
int main()
{
freopen("snow.in","r",stdin);
freopen("snow.out","w",stdout);
er[0]=1;fo(i,1,18)er[i]=er[i-1]<<1;
int q,w,e,m_;
read(n),read(m_);
fo(i,1,n-1)read(q),read(w),link(q,w);
dfs1(1,0,1);
fo(j,1,18)fo(i,1,rmq0)rmq[i][j]=rmx(i,min(rmq0,i+er[j-1]),j-1);
build(1,n,1);
while(m_--)
{
read(q),read(w);
if(a[q].zx>a[w].zx)swap(q,w);
wwqq t=merge(find(1,n,1,1,a[q].zx-1),find(1,n,1,a[q].en+1,a[w].zx-1),find(1,n,1,max(a[w].en,a[q].en)+1,n));
printf("%d\n",t.s);
}
return 0;
}