正题

题目链接:https://www.luogu.com.cn/problem/P5305

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题目大意

给一棵有根树和\(k\)，\(Q\)次询问给出\(x,y\)求

\[\sum_{i=1}^{x}dep_{LCA(i,y)}^k \]

\(1\leq n,Q\leq 5\times 10^5,1\leq k\leq 10^9\)

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解题思路

和之前\(LCA\)那题一样的思路，如果\(k\)等于\(1\)的话。加入一个点\(i\)就把\(i\)到根节点的路径上加一，然后询问节点\(x\)就查询\(x\)到根节点路径的和。

这题的话多了个\(k\)，其实是一样的，只是每个点的权值要调成\(dep_{x}^k-(dep_{x}-1)^k\)，然后询问路径上权值*点权的和就好了。

因为\(k\)是固定的，所以直接开个线段树就很好搞了。

时间复杂度\(O(n\log^2 n)\)

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code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const ll N=5e5+10,P=998244353;
struct node{
	ll to,next;
}a[N];
struct qnode{
	ll l,x,id;
}q[N];
ll n,Q,k,tot,pw[N],ls[N],fa[N],ans[N];
ll cnt,siz[N],dep[N],son[N],top[N],dfn[N],rfn[N];
ll w[N<<2],v[N<<2],lazy[N<<2];
ll power(ll x,ll b){
	ll ans=1;
	while(b){
		if(b&1)ans=ans*x%P;
		x=x*x%P;b>>=1;
	}
	return ans;
}
void addl(ll x,ll y){
	a[++tot].to=y;
	a[tot].next=ls[x];
	ls[x]=tot;return;
}
bool cmp(qnode x,qnode y)
{return x.l<y.l;}
void dfs1(ll x){
	siz[x]=1;dep[x]=dep[fa[x]]+1;
	for(ll i=ls[x];i;i=a[i].next){
		ll y=a[i].to;
		dfs1(y);siz[x]+=siz[y];
		if(siz[y]>siz[son[x]])son[x]=y;
	}
	return;
}
void dfs2(ll x){
	dfn[++cnt]=x;rfn[x]=cnt;
	if(son[x]){
		top[son[x]]=top[x];
		dfs2(son[x]);
	}
	for(ll i=ls[x];i;i=a[i].next){
		ll y=a[i].to;
		if(y==son[x])continue;
		top[y]=y;dfs2(y);
	}
	return;
}
void Build(ll x,ll l,ll r){
	if(l==r){
		w[x]=(pw[dep[dfn[l]]]-pw[dep[dfn[l]]-1]+P)%P;
		return;
	}
	ll mid=(l+r)>>1;
	Build(x*2,l,mid);Build(x*2+1,mid+1,r);
	w[x]=(w[x*2]+w[x*2+1])%P;
	return;
}
void Downdata(ll x){
	if(!lazy[x])return;
	lazy[x*2]+=lazy[x];lazy[x*2+1]+=lazy[x];
	(v[x*2]+=w[x*2]*lazy[x]%P)%=P;
	(v[x*2+1]+=w[x*2+1]*lazy[x]%P)%=P;
	lazy[x]=0;return;
}
void Change(ll x,ll L,ll R,ll l,ll r){
	if(L==l&&R==r){lazy[x]++;(v[x]+=w[x])%=P;return;}
	ll mid=(L+R)>>1;Downdata(x);
	if(r<=mid)Change(x*2,L,mid,l,r);
	else if(l>mid)Change(x*2+1,mid+1,R,l,r);
	else Change(x*2,L,mid,l,mid),Change(x*2+1,mid+1,R,mid+1,r);
	v[x]=(v[x*2]+v[x*2+1])%P;return;
}
ll Ask(ll x,ll L,ll R,ll l,ll r){
	if(L==l&&R==r)return v[x];
	ll mid=(L+R)>>1;Downdata(x);
	if(r<=mid)return Ask(x*2,L,mid,l,r);
	if(l>mid)return Ask(x*2+1,mid+1,R,l,r);
	return (Ask(x*2,L,mid,l,mid)+Ask(x*2+1,mid+1,R,mid+1,r))%P;
}
void Add(ll x){
	while(x)
		Change(1,1,n,rfn[top[x]],rfn[x]),x=fa[top[x]];
	return;
}
ll Ask(ll x){
	ll ans=0;
	while(x)
		(ans+=Ask(1,1,n,rfn[top[x]],rfn[x]))%=P,x=fa[top[x]];
	return ans;
}
signed main()
{
	scanf("%lld%lld%lld",&n,&Q,&k);
	for(ll i=1;i<=n;i++)pw[i]=power(i,k);
	for(ll i=2;i<=n;i++){
		scanf("%lld",&fa[i]);
		addl(fa[i],i);
	}
	for(ll i=1;i<=Q;i++){
		scanf("%lld%lld",&q[i].l,&q[i].x);
		q[i].id=i;
	}
	sort(q+1,q+1+Q,cmp);
	top[1]=1;dfs1(1);dfs2(1);
	Build(1,1,n);ll z=1;
	for(ll i=1;i<=n;i++){
		Add(i);
		while(z<=Q&&q[z].l<=i)
			ans[q[z].id]=Ask(q[z].x),z++;
	}
	for(ll i=1;i<=Q;i++)
		printf("%lld\n",ans[i]);
	return 0;
}

标签：lazy,return,剖分,ll,mid,树链,GZOI2019,son,top
来源： https://www.cnblogs.com/QuantAsk/p/14870892.html