cf396C. On Changing Tree
题解 操作一可以转化成 x+dep[i]*k-dep[j]*k j是i的子树节点 然后dfs序以后用两个树状数组维护即可
#include <algorithm> #include <iostream> #include <cstring> #include <cstdio> #include <vector> #include <stack> #include <queue> #include <cmath> #include <set> #include <map> #define mp make_pair #define pb push_back #define pii pair<int,int> #define link(x) for(edge *j=h[x];j;j=j->next) #define inc(i,l,r) for(int i=l;i<=r;i++) #define dec(i,r,l) for(int i=r;i>=l;i--) const int MAXN=3e5+10; const double eps=1e-8; #define ll long long const int mod=1e9+7; using namespace std; struct edge{int t;edge*next;}e[MAXN],*h[MAXN],*o=e; void add(int x,int y){o->t=y;o->next=h[x];h[x]=o++;} ll read(){ ll x=0,f=1;char ch=getchar(); while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();} while(isdigit(ch))x=x*10+ch-'0',ch=getchar(); return x*f; } ll sum1[MAXN],sum2[MAXN]; int p[MAXN],cnt,dep[MAXN],num[MAXN]; void dfs(int x,int fa,int deep){ p[x]=++cnt;dep[x]=deep+1;num[x]=1; link(x){ if(j->t!=fa){ dfs(j->t,x,deep+1); num[x]+=num[j->t]; } } } int n; int get_id(int x){return x&(-x);} void update1(int x,ll k){ for(int i=x;i<=n+1;i+=get_id(i))sum1[i]+=k,sum1[i]%=mod; } void update2(int x,ll k){ for(int i=x;i<=n+1;i+=get_id(i))sum2[i]+=k,sum2[i]%=mod; } ll Sum1(int x){ ll ans=0; for(int i=x;i>0;i-=get_id(i))ans+=sum1[i],ans%=mod; return ans; } ll Sum2(int x){ ll ans=0; for(int i=x;i>0;i-=get_id(i))ans+=sum2[i],ans%=mod; return ans; } int main(){ n=read(); int u,v,op;ll x,k; inc(i,2,n)u=read(),add(u,i); dfs(1,0,0); int q=read(); while(q--){ op=read(); if(op==1){ v=read();x=read();k=read(); ll t1=(x+(1ll*dep[v]*k%mod))%mod; //cout<<t1<<endl; update1(p[v],t1); update1(p[v]+num[v],(mod-t1)); update2(p[v],k); update2(p[v]+num[v],mod-k); } else{ v=read(); ll ans=Sum1(p[v]); //cout<<ans<<"==="<<endl; ll ans2=Sum2(p[v]); ans2=ans2*dep[v]%mod; ans=(ans-ans2+mod)%mod; printf("%lld\n",ans); } } return 0; }
You are given a rooted tree consisting of n vertices numbered from 1 to n. The root of the tree is a vertex number 1.
Initially all vertices contain number 0. Then come q queries, each query has one of the two types:
- The format of the query: 1 v x k. In response to the query, you need to add to the number at vertex v number x; to the numbers at the descendants of vertex v at distance 1, add x - k; and so on, to the numbers written in the descendants of vertex v at distance i, you need to add x - (i·k). The distance between two vertices is the number of edges in the shortest path between these vertices.
- The format of the query: 2 v. In reply to the query you should print the number written in vertex v modulo 1000000007 (109 + 7).
Process the queries given in the input.
The first line contains integer n (1 ≤ n ≤ 3·105) — the number of vertices in the tree. The second line contains n - 1 integers p2, p3, ... pn(1 ≤ pi < i), where pi is the number of the vertex that is the parent of vertex i in the tree.
The third line contains integer q (1 ≤ q ≤ 3·105) — the number of queries. Next q lines contain the queries, one per line. The first number in the line is type. It represents the type of the query. If type = 1, then next follow space-separated integers v, x, k (1 ≤ v ≤ n; 0 ≤ x < 109 + 7; 0 ≤ k < 109 + 7). If type = 2, then next follows integer v (1 ≤ v ≤ n) — the vertex where you need to find the value of the number.
For each query of the second type print on a single line the number written in the vertex from the query. Print the number modulo 1000000007 (109 + 7).
3
1 1
3
1 1 2 1
2 1
2 2
2
1
You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).