【JZOJ 4604】【BZOJ 4551】树

Description

Solution

一道树链剖分裸题，
复杂度：$O(n\log_2(n)^2)$

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[N],B0=1;
struct qqww
{int s,lt,fa,zx;}a[N];
int zx[N],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;
}