洛谷$P4211 [LNOI2014] LCA$ 树链剖分+线段树

正解:树剖+线段树

解题报告:

传送门$QwQ$

看到$dep[lca]$啥的就想到之前托腮腮$CSP$模拟$D1T3$的那个套路,,,

然后试下这个想法,于是$dep[lca(x,y)]=sum_{i=1}^{infty}[ileq dep[lca(x,y)]]$,就可以是,从$x$到根全部加一然后查询$y$到根的权值和.

现在变成$sum_{i=l}^r dep[lca(i,x)]$,那就$l$到$r$到根全加一然后查询$x$到根的权值和.

显然考虑差分呗?就$1$到$r$全加一的权值和减去$1$到$l-1$全加一的权值和.

于是就从$1$枚举到$n$每次从当前节点到根全加一,顺便维护下这个答案就完事$QwQ$

 

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define lowbit(x) (x&(-x))
#define rg register
#define gc getchar()
#define ls(x) (x<<1)
#define rs(x) ((x<<1)|1)
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define my(i,x,y) for(ri i=x;i>=y;--i)
#define e(i,x) for(ri i=head[x];~i;i=edge[i].nxt)

const int N=50000+5,mod=201314;
int n,q,dfn[N],sz[N],top[N],fa[N],as[N],hs[N],dfn_cnt,tr[N<<2],ad[N<<2];
vector<int>son[N];
struct node{int id,x,op;};
vector<node>V[N];

il int read()
{
    rc ch=gc;ri x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
il void add(ri &x,ri y){x+=y;if(x>=mod)x-=mod;if(x<0)x+=mod;}
il void dfs1(ri x)
{
    sz[x]=1;ri siz=son[x].size();
    rp(i,0,siz-1){dfs1(son[x][i]),sz[x]+=sz[son[x][i]];if(sz[son[x][i]]>sz[hs[x]])hs[x]=son[x][i];}
}
il void dfs2(ri x,ri tp)
{
    top[x]=tp;dfn[x]=++dfn_cnt;if(hs[x])dfs2(hs[x],tp);ri siz=son[x].size();
    rp(i,0,siz-1)if(son[x][i]^hs[x])dfs2(son[x][i],son[x][i]);
}
il void pushdown(ri x,ri l,ri r)
{
    if(!ad[x])return;
    ri mid=(l+r)>>1;
    add(tr[ls(x)],1ll*ad[x]*(mid-l+1)%mod),add(tr[rs(x)],1ll*ad[x]*(r-mid)%mod);
    ad[ls(x)]+=ad[x],ad[rs(x)]+=ad[x];ad[x]=0;
}
void modify(ri x,ri l,ri r,ri to_l,ri to_r)
{
    if(to_l<=l && r<=to_r){add(tr[x],r-l+1);add(ad[x],1);return;}
    pushdown(x,l,r);
    ri mid=(l+r)>>1;if(mid>=to_l)modify(ls(x),l,mid,to_l,to_r);if(mid<to_r)modify(rs(x),mid+1,r,to_l,to_r);
    tr[x]=(tr[ls(x)]+tr[rs(x)])%mod;
}
int query(ri x,ri l,ri r,ri to_l,ri to_r)
{
    if(to_l<=l && r<=to_r)return tr[x];
    pushdown(x,l,r);ri mid=(l+r)>>1,ret=0;
    if(mid>=to_l)ret=query(ls(x),l,mid,to_l,to_r);;if(mid<to_r)add(ret,query(rs(x),mid+1,r,to_l,to_r));
    return ret;
}

int main()
{
    freopen("4211.in","r",stdin);freopen("4211.out","w",stdout);
    n=read();q=read();rp(i,2,n)son[fa[i]=read()+1].push_back(i);dfs1(1);dfs2(1,1);
    rp(i,1,q){ri l=read(),r=read()+1,x=read()+1;V[l].push_back((node){i,x,-1});V[r].push_back((node){i,x,1});}
    rp(i,1,n)
    {
        ri nw=i;while(nw)modify(1,1,n,dfn[top[nw]],dfn[nw]),nw=fa[top[nw]];
        for(auto j:V[i])
        {nw=j.x;while(nw)add(as[j.id],j.op*query(1,1,n,dfn[top[nw]],dfn[nw])),nw=fa[top[nw]];}
    }
    rp(i,1,q)printf("%d
",as[i]);
    return 0;
}