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
$$\sum _{i=1}^{x}de{p}_{LCA(i,y)}^k$$

$1\le n,Q\le 5\times 10^5,1\le k\le 10^9$

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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$de{p}_x^k-(de{p}_x-1{)}^k$，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;
}