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【JZOJ 4604】【BZOJ 4551】树

2016-07-11 16:52 204 查看

Description



Solution

一道树链剖分裸题,

复杂度:O(nlog2(n)2)

Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#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)
using namespace std;
const int N=100500,maxlongint=2107483640;
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
,B0=1;
struct qqww
{int s,lt,fa,zx;}a
;
int zx
,b[N*3];
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 dfs1(int q,int fa)
{
a[q].fa=fa;a[q].s=1;
efo(i,q)iff()
a[q].s+=dfs1(B[i][1],q);
return a[q].s;
}
void dfs2(int q,int fa,int lt)
{
if(!lt)lt=q;a[q].lt=lt;zx[a[q].zx=++zx[0]]=q;
int mx=0;
efo(i,q)iff()if(a[mx].s<a[B[i][1]].s)mx=B[i][1];
if(mx)dfs2(mx,q,lt);
efo(i,q)iff()if(B[i][1]!=mx)dfs2(B[i][1],q,0);
}
void change(int l,int r,int e,int l1)
{
if(l==r){b[e]=l;return;}
int t=(l+r)/2;
if(l1<=t)change(l,t,e*2,l1);
else change(t+1,r,e*2+1,l1);
if(b[e*2+1])b[e]=b[e*2+1];
else b[e]=b[e*2];
}
int find(int l,int r,int e,int l1,int r1)
{
if(l1<=l&&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
{
int s=find(t+1,r,e*2+1,t,r1);
if(s)return s;
return find(l,t,e*2,l1,t);
}
}
int search(int q)
{
while(q)
{
int t=a[q].lt;
int s=find(1,n,1,a[t].zx,a[q].zx);
if(s)return zx[s];
q=a[t].fa;
}
return 1;
}
int main()
{
int q,w,m_;
read(n),read(m_);
fo(i,1,n-1)read(q),read(w),link(q,w);
dfs1(1,0);
dfs2(1,0,0);
change(1,n,1,1);
while(m_--)
{
char ch=' ';while(ch!='Q'&&ch!='C')ch=getchar();
read(q);
if(ch=='C')change(1,n,1,a[q].zx);
else printf("%d\n",search(q));
}
return  0;
}
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