树上操作 HYSBZ - 4034

https://www.lydsy.com/JudgeOnline/problem.php?id=4034

在树链剖分后形成的dfs序 和普通的dfs序性质一样 一颗子树在dfs序列中仍然是连续的 因为树剖之后虽然先走重链 但还是要走完当前子树后才会递归返回 然后进入其他子树

#include <bits/stdc++.h>
using namespace std;
#define ll long long

struct node1
{
    int v;
    int next;
};

struct node2
{
    int l;
    int r;
    ll val;
    ll laz;
};

node1 edge[200010];
node2 tree[400010];
int pre[100010],first[100010],fa[100010],deep[100010],sum[100010],son[100010],top[100010],mp1[100010],mp2[100010];
int n,q,num;

void addedge(int u,int v)
{
    edge[num].v=v;
    edge[num].next=first[u];
    first[u]=num++;
}

void dfsI(int cur)
{
    int i,v;
    sum[cur]=1,son[cur]=-1;
    for(i=first[cur];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(v!=fa[cur])
        {
            fa[v]=cur,deep[v]=deep[cur]+1;
            dfsI(v);
            sum[cur]+=sum[v];
            if(son[cur]==-1||sum[son[cur]]<sum[v])
            {
                son[cur]=v;
            }
        }
    }
}

void dfsII(int cur,int tp)
{
    int i,v;
    num++;
    top[cur]=tp,mp1[cur]=num,mp2[num]=cur;
    if(son[cur]==-1) return;
    dfsII(son[cur],tp);
    for(i=first[cur];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(v!=fa[cur]&&v!=son[cur])
        {
            dfsII(v,v);
        }
    }
}

void pushup(int cur)
{
    tree[cur].val=tree[2*cur].val+tree[2*cur+1].val;
    return;
}

void pushdown(int cur)
{
    if(tree[cur].laz!=0)
    {
        tree[2*cur].val+=(tree[2*cur].r-tree[2*cur].l+1)*tree[cur].laz;
        tree[2*cur].laz+=tree[cur].laz;
        tree[2*cur+1].val+=(tree[2*cur+1].r-tree[2*cur+1].l+1)*tree[cur].laz;
        tree[2*cur+1].laz+=tree[cur].laz;
        tree[cur].laz=0;
    }
    return;
}

void build(int l,int r,int cur)
{
    int m;
    tree[cur].l=l;
    tree[cur].r=r;
    tree[cur].laz=0;
    if(l==r)
    {
        tree[cur].val=pre[mp2[l]];
        return;
    }
    m=(l+r)/2;
    build(l,m,2*cur);
    build(m+1,r,2*cur+1);
    pushup(cur);
}

void updateI(int tar,ll val,int cur)
{
    if(tree[cur].l==tree[cur].r)
    {
        tree[cur].val+=val;
        return;
    }
    pushdown(cur);
    if(tar<=tree[2*cur].r) updateI(tar,val,2*cur);
    else updateI(tar,val,2*cur+1);
    pushup(cur);
}

void updateII(int pl,int pr,ll val,int cur)
{
    if(pl<=tree[cur].l&&tree[cur].r<=pr)
    {
        tree[cur].val+=(tree[cur].r-tree[cur].l+1)*val;
        tree[cur].laz+=val;
        return;
    }
    pushdown(cur);
    if(pl<=tree[2*cur].r) updateII(pl,pr,val,2*cur);
    if(pr>=tree[2*cur+1].l) updateII(pl,pr,val,2*cur+1);
    pushup(cur);
    return;
}

ll query(int pl,int pr,int cur)
{
    ll res;
    if(pl<=tree[cur].l&&tree[cur].r<=pr)
    {
        return tree[cur].val;
    }
    pushdown(cur);
    res=0;
    if(pl<=tree[2*cur].r) res+=query(pl,pr,2*cur);
    if(pr>=tree[2*cur+1].l) res+=query(pl,pr,2*cur+1);
    return res;
}

ll solve(int u,int v)
{
    ll res;
    res=0;
    while(top[u]!=top[v])
    {
        if(deep[top[u]]<deep[top[v]]) swap(u,v);
        res+=query(mp1[top[u]],mp1[u],1);
        u=fa[top[u]];
    }
    if(deep[u]<deep[v]) swap(u,v);
    res+=query(mp1[v],mp1[u],1);
    return res;
}

int main()
{
    ll w;
    int i,u,v,op;
    while(scanf("%d%d",&n,&q)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&pre[i]);
        }
        memset(first,-1,sizeof(first));
        num=0;
        for(i=1;i<=n-1;i++)
        {
            scanf("%d%d",&u,&v);
            addedge(u,v);
            addedge(v,u);
        }
        fa[1]=-1,deep[1]=1;
        dfsI(1);
        num=0;
        dfsII(1,1);
        build(1,n,1);
        while(q--)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%lld",&u,&w);
                updateI(mp1[u],w,1);
            }
            else if(op==2)
            {
                scanf("%d%lld",&u,&w);
                updateII(mp1[u],mp1[u]+sum[u]-1,w,1);
            }
            else
            {
                scanf("%d",&u);
                printf("%lld\n",solve(1,u));
            }
        }
    }
    return 0;
}