BZOJ 3626 LCA

我们考虑一个暴力：若两点求题目中所需距离，那么选一个点到根的路径上所有点权+1。然后求另一个点到根的点权和即可。

然后？这个操作是可加的，可减的，因此可以具备前缀操作的性质。

把所有询问离线，然后链剖。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxv 100500
#define maxq 200500
#define maxe 100500
#define mod 201314
using namespace std;
struct edge
{
    long long v,nxt;
}e[maxe];
struct aasskk
{
    long long id,pos,r,sgn;
}ask[maxq];
long long n,q,x,y,z,g[maxv],top[maxv],fath[maxv],son[maxv],dis[maxv],w[maxv],size[maxv];
long long ls[maxv<<2],rs[maxv<<2],value[maxv<<2],lazy[maxv<<2],nume=0,numq=0,cnt=0,tot=0;
long long ans[maxq],root;
bool cmp(aasskk x,aasskk y)
{
    return x.pos<y.pos;
}
void addedge(long long u,long long v)
{
    e[++nume].v=v;
    e[nume].nxt=g[u];
    g[u]=nume;
}
void addask(long long id,long long pos,long long r,long long sgn)
{
    ask[++numq].id=id;
    ask[numq].pos=pos;
    ask[numq].r=r;
    ask[numq].sgn=sgn;
}
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])
        {
            fath[v]=x;dis[v]=dis[x]+1;
            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;top[x]=father;
    if (son[x]!=0) dfs2(son[x],father);
    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);
    }
}
void build(long long &now,long long left,long long right)
{
    now=++tot;
    value[now]=0;lazy[now]=0;
    if (left==right) return;
    long long mid=(left+right)>>1;
    build(ls[now],left,mid);
    build(rs[now],mid+1,right);
}
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];
        lazy[rs[now]]+=lazy[now];
        value[ls[now]]+=(mid-left+1)*lazy[now];
        value[rs[now]]+=(right-mid)*lazy[now];
        lazy[now]=0;
    }
}
void big_modify(long long now,long long left,long long right,long long l,long long r)
{
    pushdown(now,left,right);
    if ((left==l) && (right==r))
    {
        lazy[now]+=1;
        value[now]+=(right-left+1);
        return;
    }
    long long mid=(left+right)>>1;
    if (r<=mid) big_modify(ls[now],left,mid,l,r);
    else if (l>=mid+1) big_modify(rs[now],mid+1,right,l,r);
    else
    {
        big_modify(ls[now],left,mid,l,mid);
        big_modify(rs[now],mid+1,right,mid+1,r);
    }
    value[now]=value[ls[now]]+value[rs[now]];
}
long long query(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 query(ls[now],left,mid,l,r);
    else if (l>=mid+1) return query(rs[now],mid+1,right,l,r);
    else return query(ls[now],left,mid,l,mid)+query(rs[now],mid+1,right,mid+1,r);
}
void work1(long long x)
{
    long long f1=top[x];
    while (f1!=1)
    {
        big_modify(root,1,cnt,w[f1],w[x]);
        x=fath[f1];f1=top[x];
    }
    big_modify(root,1,cnt,w[1],w[x]);
}
void work2(long long x)
{
    long long r=ask[x].r,ret=0;
    long long f1=top[r];
    while (f1!=1)
    {
        ret=(ret+query(root,1,cnt,w[f1],w[r]))%mod;
        r=fath[f1];f1=top[r];
    }
    ret=(ret+query(root,1,cnt,w[1],w[r]))%mod;
    ans[ask[x].id]=(ans[ask[x].id]+ret*ask[x].sgn)%mod;
    ans[ask[x].id]=(ans[ask[x].id]+mod)%mod;
}
int main()
{
    scanf("%lld%lld",&n,&q);
    for (long long i=1;i<=n-1;i++)
    {
        scanf("%lld",&x);
        addedge(x+1,i+1);
        addedge(i+1,x+1);
    }
    for (long long i=1;i<=q;i++)
    {
        scanf("%lld%lld%lld",&x,&y,&z);
        if (x!=0) addask(i,x,z+1,-1);
        addask(i,y+1,z+1,1);
    }
    sort(ask+1,ask+numq+1,cmp);
    dfs1(1);
    dfs2(1,1);
    build(root,1,cnt);
    long long now=1;
    for (long long i=1;i<=n;i++)
    {
        work1(i);
        while (ask[now].pos==i)
        {
            work2(now);
            now++;
        }
    }
    for (long long i=1;i<=q;i++)
    {
        while (ans[i]<0) ans[i]=(ans[i]+mod)%mod;
        printf("%lld\n",ans[i]%mod);
    }
    return 0;
}