[jzoj 5662] 尺树寸泓 解题报告 (线段树+中序遍历)

interlinkage:

https://jzoj.net/senior/#contest/show/2703/1

description:

solution:

• 发现$dfs$序不好维护
• 注意到这是一棵平衡树，左旋右旋中序遍历不会改变，且一个点的子树在中序遍历上也是一个连续的区间
• 每次旋转只改变两个点的力量值
• 在中序遍历上建线段树维护单点修改,区间乘积查询就好

code:

#include<algorithm>
#include<cstring>
#include<iostream>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
typedef long long ll;

const int N=2e5+;
const ll mo=1e9+;
int n,q,root,tim;
int t[N][],fa[N],id[N],dfn[N],size[N];
ll w[N],mul[N<<],sum[N];
inline ll read()
{
    char ch=getchar();ll s=,f=;
    while (ch<''||ch>'') {if (ch=='-') f=-;ch=getchar();}
    while (ch>=''&&ch<='') {s=(s<<)+(s<<)+ch-'';ch=getchar();}
    return s*f;
}
void up1(int o)
{
    size[o]=size[t[o][]]+size[t[o][]]+;
    sum[o]=(sum[t[o][]]+sum[t[o][]]+w[o])%mo;
}
void dfs(int x)
{
    if (t[x][]) dfs(t[x][]);
    dfn[++tim]=x;id[x]=tim;
    if (t[x][]) dfs(t[x][]);
    up1(x);
}
void up2(int o)
{
    mul[o]=mul[o<<]*mul[o<<|]%mo;
}
void build(int o,int l,int r)
{
    if (l==r)
    {
        mul[o]=sum[dfn[l]];
        return;
    }
    int mid=l+r>>;
    build(o<<,l,mid);
    build(o<<|,mid+,r);
    up2(o);
}
void rotate(int x,int op)
{
    int y=fa[x],z=fa[y];
    if (z) t[z][t[z][]==y]=x;
    fa[x]=z;
    t[y][op]=t[x][!op];fa[t[x][!op]]=y;
    t[x][!op]=y;fa[y]=x;
    up1(y);up1(x);
}
void ins(int o,int l,int r,int pos)
{
    if (l==r)
    {
        mul[o]=sum[dfn[pos]];
        return;
    }
    int mid=l+r>>;
    if (pos<=mid) ins(o<<,l,mid,pos);
    else ins(o<<|,mid+,r,pos);
    up2(o);
}
ll query(int o,int l,int r,int x,int y)
{
    if (l>=x&&r<=y) return mul[o];
    int mid=l+r>>;
    ll re=;
    if (x<=mid) re=re*query(o<<,l,mid,x,y)%mo;
    if (y>mid) re=re*query(o<<|,mid+,r,x,y)%mo;
    return re;
}
int main()
{
    freopen("splay.in","r",stdin);
    freopen("splay.out","w",stdout);
    n=read();q=read();
    for (int i=;i<=n;i++)
    {
        w[i]=read();t[i][]=read();t[i][]=read();
        if (t[i][]) fa[t[i][]]=i;
        if (t[i][]) fa[t[i][]]=i;
    }
    for (int i=;i<=n;i++) if (!fa[i]) root=i;
    dfs();
    build(,,tim);
    while (q--)
    {
        int op=read(),x=read();
        if (op<=)
        {
            if (t[x][op]) x=t[x][op];
            else continue;
            rotate(x,op);
            ins(,,tim,id[x]);
            if (t[x][!op]) ins(,,tim,id[t[x][!op]]);
        }
        if (op==)
        {
            printf("%lld\n",query(,,tim,id[x]-size[t[x][]],id[x]+size[t[x][]]));
        }
    }
    return ;
}