BZOJ 4034 T2

树剖。观察这样一个事实：一个子树上的区间一定是连续的。打lazy标记即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#define maxv 200500
#define maxe 200500
using namespace std;
long long n,m,x,y,type,nume=0;
long long dis[maxv],fath[maxv],son[maxv],w[maxv],mx[maxv],size[maxv],top[maxv],fw[maxv],k[maxv],g[maxv];
long long cnt=0,tot=0,ls[maxv<<3],rs[maxv<<3],value[maxv<<3],lazy[maxv<<3],root;
struct edge
{
    long long v,nxt;
}e[maxe];
void addedge(long long u,long long v)
{
    e[++nume].v=v;
    e[nume].nxt=g[u];
    g[u]=nume;
}
void dfs1(long long x)
{
    size[x]=1;son[x]=0;
    for (long long i=g[x];i;i=e[i].nxt)
    {
        long long v=e[i].v;
        if (v!=fath[x])
        {
            dis[v]=dis[x]+1;
            fath[v]=x;
            dfs1(v);
            size[x]=size[x]+size[v];
            if (size[v]>size[son[x]]) son[x]=v;
        }
    }    
}
void dfs2(long long x,long long father)
{
    w[x]=++cnt;fw[cnt]=x;top[x]=father;mx[x]=w[x];
    if (son[x]!=0) dfs2(son[x],father);mx[x]=max(mx[x],mx[son[x]]);
    for (long long i=g[x];i;i=e[i].nxt)
    {
        long long v=e[i].v;
        if ((v!=fath[x]) && (v!=son[x]))
        {
            dfs2(v,v);
            mx[x]=max(mx[x],mx[v]);
        }
    }
}
void build(long long &now,long long left,long long right)
{
    now=++tot;value[now]=0;lazy[now]=0;
    if (left==right) 
    {
        value[now]=k[fw[left]];
        return;
    }
    long long mid=(left+right)>>1;
    build(ls[now],left,mid);
    build(rs[now],mid+1,right);
    value[now]=value[ls[now]]+value[rs[now]];
}
void pushdown(long long now,long long left,long long right)
{
    if (lazy[now]!=0)
    {
        long long mid=(left+right)>>1;
        lazy[ls[now]]+=lazy[now];
        value[ls[now]]+=lazy[now]*(mid-left+1);
        lazy[rs[now]]+=lazy[now];
        value[rs[now]]+=lazy[now]*(right-mid);
        lazy[now]=0;
    }
}
void single_modify(long long now,long long left,long long right,long long pos,long long a)
{
    pushdown(now,left,right);
    if ((left==right) && (left==pos))
    {
        value[now]+=a;
        return;
    }
    long long mid=(left+right)>>1;
    if (pos<=mid) single_modify(ls[now],left,mid,pos,a);
    else single_modify(rs[now],mid+1,right,pos,a);
    value[now]=value[ls[now]]+value[rs[now]];
}
void big_modify(long long now,long long left,long long right,long long l,long long r,long long a)
{
    pushdown(now,left,right);
    if ((left==l) && (right==r))
    {
        value[now]+=a*(right-left+1);
        lazy[now]+=a;
        return;
    }
    long long mid=(left+right)>>1;
    if (r<=mid) big_modify(ls[now],left,mid,l,r,a);
    else if (l>=mid+1) big_modify(rs[now],mid+1,right,l,r,a);
    else
    {
        big_modify(ls[now],left,mid,l,mid,a);
        big_modify(rs[now],mid+1,right,mid+1,r,a);
    }
    value[now]=value[ls[now]]+value[rs[now]];
}
long long ask(long long now,long long left,long long right,long long l,long long r)
{
    pushdown(now,left,right);
    if ((left==l) && (right==r))
        return value[now];
    long long mid=(left+right)>>1;
    if (r<=mid) return ask(ls[now],left,mid,l,r);
    else if (l>=mid+1) return ask(rs[now],mid+1,right,l,r);
    else return ask(ls[now],left,mid,l,mid)+ask(rs[now],mid+1,right,mid+1,r);
}
void work1()
{
    long long a;
    scanf("%lld%lld",&x,&a);
    single_modify(root,1,cnt,w[x],a);
}
void work2()
{
    long long a;
    scanf("%lld%lld",&x,&a);
    big_modify(root,1,cnt,w[x],mx[x],a);
}
void work3()
{
    long long ans=0;
    scanf("%lld",&x);
    long long f1=top[x];
    while (f1!=1)
    {
        ans=ans+ask(root,1,cnt,w[f1],w[x]);
        x=fath[f1];
        f1=top[x];
    }
    ans=ans+ask(root,1,cnt,w[1],w[x]);
    printf("%lld\n",ans);
}
int main()
{
    scanf("%lld%lld",&n,&m);
    for (long long i=1;i<=n;i++)
        scanf("%lld",&k[i]);
    for (long long i=1;i<=n-1;i++)
    {
        scanf("%lld%lld",&x,&y);
        addedge(x,y);
        addedge(y,x);
    }
    dfs1(1);
    dfs2(1,1);
    build(root,1,cnt);
    for (long long i=1;i<=m;i++)
    {
        scanf("%lld",&type);
        if (type==1) work1();
        else if (type==2) work2();
        else work3();
    }
    return 0;
}