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2020-2021:teams:wangzai_milk:树链剖分 [CVBB ACM Team]

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2020-2021:teams:wangzai_milk:树链剖分

这是本文档旧的修订版!


树链剖分

是什么

怎么用

普通用法

trick | keys:LCA

感觉这两道题有一点点共性但又不是显式的,还是放在一起了,无论如何挺值得两连做的。建议思考,仿佛有智力开发的效果

[BZOJ 3626][LNOI2014]LCA

给出$n$个点的有根树,$q$次询问,每次询问给出$l,r,z$,求$\sum\limits_{l\le i\le r}\text{dep}[\text{LCA}(i,z)]$。$1\le n,q\le50000$

题解:定义询问$(x,z)$为$\sum\limits_{1\le i\le x}\text{dep}[\text{LCA}(i,z)]$,则可以将题中询问拆为询问$(r,z)$和$(l-1,z)$的差。于是可以离线,从$1$到$n$逐个加入,加入时将该节点到根节点路径上点的值各加一,线段树维护。而对于每个$z$可以在加入$x$刚好完成时查询$z$到根节点路径上点的值来得到答案。

code:

点击以显示 ⇲

点击以隐藏 ⇱

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define MAXN 50005
#define Mod 201314
using namespace std;
int n,q,deep[MAXN],head[MAXN],cnt=0,tot=0;
int sz=0,maxv[MAXN],father[MAXN],siz[MAXN],top[MAXN],pos[MAXN],res[MAXN*2];
struct Node1
{
    int next,to;
}Edges[MAXN*2];
struct Node2
{
    int l,r,sum,lazy;
}t[MAXN*4];
struct Node3
{
    int x,z,id;
    Node3(int x=0,int z=0,int id=0):x(x),z(z),id(id){}
}Q[MAXN*2];
bool cmp(Node3 a,Node3 b){return a.x<b.x;}
int read()
{
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
void addedge(int u,int v)
{
    Edges[++cnt].next=head[u];
    head[u]=cnt;
    Edges[cnt].to=v;
}
void dfs1(int u)
{
    siz[u]=1;int k=-1;
    for(int i=head[u];~i;i=Edges[i].next)
    {
        int v=Edges[i].to;
        deep[v]=deep[u]+1,dfs1(v);
        siz[u]+=siz[v];
        if(k==-1||siz[v]>siz[k])k=v;
    }
    maxv[u]=k;
}
void dfs2(int u,int t)
{
    ++sz,pos[u]=sz;
    top[u]=t;
    if(maxv[u]!=-1)dfs2(maxv[u],t);
    for(int i=head[u];~i;i=Edges[i].next)
    if(maxv[u]!=Edges[i].to)dfs2(Edges[i].to,Edges[i].to);
}
void build(int idx,int l,int r)
{
    t[idx].l=l,t[idx].r=r;
    if(l==r)return;
    int mid=(l+r)>>1;
    build(idx<<1,l,mid),build(idx<<1|1,mid+1,r);
}
void pushdown(int idx)
{
    if(t[idx].l<t[idx].r&&t[idx].lazy)
    {
        t[idx<<1].lazy+=t[idx].lazy;
        t[idx<<1|1].lazy+=t[idx].lazy;
        t[idx<<1].sum+=t[idx].lazy*(t[idx<<1].r-t[idx<<1].l+1);
        t[idx<<1].sum%=Mod;
        t[idx<<1|1].sum+=t[idx].lazy*(t[idx<<1|1].r-t[idx<<1|1].l+1);
        t[idx<<1|1].sum%=Mod;
        t[idx].lazy=0;
    }
}
void add(int idx,int l,int r)
{
    if(l<=t[idx].l&&r>=t[idx].r)
    {
        t[idx].sum+=t[idx].r-t[idx].l+1;
        t[idx].lazy++;return;
    }
    pushdown(idx);
    int mid=(t[idx].l+t[idx].r)>>1;
    if(r<=mid)add(idx<<1,l,r);
    else if(l>mid)add(idx<<1|1,l,r);
    else add(idx<<1,l,r),add(idx<<1|1,l,r);
    t[idx].sum=(t[idx<<1].sum+t[idx<<1|1].sum)%Mod;
}
int query(int idx,int l,int r)
{
    if(l<=t[idx].l&&r>=t[idx].r)return t[idx].sum;
    pushdown(idx);
    int mid=(t[idx].l+t[idx].r)>>1;
    if(r<=mid)return query(idx<<1,l,r);
    else if(l>mid)return query(idx<<1|1,l,r);
    else return (query(idx<<1,l,r)+query(idx<<1|1,l,r))%Mod;
}
void change(int x)
{
    while(top[x])
    {
        add(1,pos[top[x]],pos[x]);
        x=father[top[x]];
    }
    add(1,pos[top[x]],pos[x]);
}
int ask(int x)
{
    int res=0;
    while(top[x])
    {
        res=(res+query(1,pos[top[x]],pos[x]))%Mod;
        x=father[top[x]];
    }
    res=(res+query(1,pos[top[x]],pos[x]))%Mod;
    return res;
}
int main()
{
    memset(head,-1,sizeof(head));
    n=read(),q=read();
    for(int i=1;i<n;i++){father[i]=read();addedge(father[i],i);}
    deep[0]=1,dfs1(0);
    dfs2(0,0);build(1,1,sz);
    for(int i=1;i<=q;i++)
    {
        int l=read(),r=read(),z=read();
        ++tot,Q[tot]=Node3(l-1,z,tot);
        ++tot,Q[tot]=Node3(r,z,tot);
    }
    sort(Q+1,Q+1+tot,cmp);
    int now=0;
    for(int i=1;i<=tot;i++)
    {
        while(now<=Q[i].x)change(now),++now;
        res[Q[i].id]=ask(Q[i].z);
    }
    for(int i=1;i<=tot;i+=2)
    printf("%d\n",(res[i+1]-res[i]+Mod)%Mod);
    return 0;
}


[BZOJ 4012][HNOI2015]开店

给出$n$个点的带权树,节点$i$的妖怪有年龄$x_i$。$Q$个询问,给出$u,l,r$询问年龄在$[l,r]$的妖怪到$u$点的距离之和,强制在线。$n\le150000,Q\le200000$,年龄上限$A\le10^9$。

题解:首先点$u,v$间的距离可以转换为$\text{dis}(u,\text{root})+\text{dis}(v,\text{root})-\text{dis}(\text{lca}(u,v),\text{root})$

2020-2021/teams/wangzai_milk/树链剖分.1590337629.txt.gz · 最后更改: 2020/05/25 00:27 由 zars19