FZU 2277 树状数组

题目链接：http://acm.fzu.edu.cn/problem.php?pid=2277

 Problem Description

There is a rooted tree with n nodes, number from 1-n. Root’s number is 1.Each node has a value ai.

Initially all the node’s value is 0.

We have q operations. There are two kinds of operations.

1 v x k : a[v]+=x , a[v’]+=x-k (v’ is child of v) , a[v’’]+=x-2*k (v’’ is child of v’) and so on.

2 v : Output a[v] mod 1000000007(10^9 + 7).

 Input

First line contains an integer T (1 ≤ T ≤ 3), represents there are T test cases.

In each test case:

The first line contains a number n.

The second line contains n-1 number, p2,p3,…,pn . pi is the father of i.

The third line contains a number q.

Next q lines, each line contains an operation. (“1 v x k” or “2 v”)

1 ≤ n ≤ 3*10^5

1 ≤ pi < i

1 ≤ q ≤ 3*10^5

1 ≤ v ≤ n; 0 ≤ x < 10^9 + 7; 0 ≤ k < 10^9 + 7

 Output

For each operation 2, outputs the answer.

 Sample Input

1 3 1 1 3 1 1 2 1 2 1 2 2

 Sample Output

2 1

 Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

题意：略

题目分析：略

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#define Pii pair<int,int>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define maxn 300100
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Mod 1000000007
using namespace std;
vector<int> G[maxn];
int n,k,x,ti=0;
LL tot[maxn];
LL C[maxn],add[maxn],Add[maxn*4];
int l[maxn],r[maxn];
LL deep[maxn];
inline int lowbit(int x){return (x&-x);}
inline void update(int x,LL y,LL t){
    while(x<=n){
        C[x]+=y;
        tot[x]+=t;
          x+=lowbit(x);
      }
}
LL query(int x){
      LL ans=0;
     while (x>0){    
        ans+=C[x];
        x-=lowbit(x);
      }
      return ans;
}
LL Query(int x){
      LL ans=0;
     while (x>0){    
        ans+=tot[x];
        x-=lowbit(x);
      }
      return ans;
}
void dfs(int x,int fa,int dep){
    l[x]=++ti;
    deep[l[x]]=dep;
    for(int i=0;i<G[x].size();i++){
        if(G[x][i]==fa) continue;
        dfs(G[x][i],x,dep+1);
    }
    r[x]=ti;
}
int main(){
    int u,Q,v,T,opt;
    cin>>T;
    while(T--){
        cin>>n;
        memset(C,0,sizeof(C));
        memset(tot,0,sizeof(tot));
        for(int i=2;i<=n;i++){
            scanf("%d",&u);
            G[u].pb(i);
        }
        ti=0;
        dfs(1,-1,1);
        LL res;
        cin>>Q;
        for(int i=1;i<=Q;i++){
            scanf("%d",&opt);
            if(opt==1){
                scanf("%d%d%d",&v,&x,&k);
                update(l[v],x+(LL)k*deep[l[v]],k);
                update(r[v]+1,-x-(LL)k*deep[l[v]],-k);
            }
            else{
                scanf("%d",&v);
                res=query(l[v])-Query(l[v])*deep[l[v]];
                while(res<0) res+=Mod;
                res%=Mod;
                printf("%I64d
",res);
            }
        }    
        for(int i=1;i<=n;i++) G[i].clear();
    }
    return 0;    
}