CodeForces 396C On Changing Tree

On Changing Tree

Time Limit: 2000ms

Memory Limit: 262144KB

This problem will be judged on CodeForces. Original ID: 396C
64-bit integer IO format: %I64d      Java class name: (Any)

 

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, addx - 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.

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.

 

Output

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).

 

Sample Input

Input

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

Output

2
1

Hint

You can read about a rooted tree here: http://en.wikipedia.org/wiki/Tree_(graph_theory).

 

Source

Codeforces Round #232 (Div. 1)

 

解题：树状数组或者线段树

 

给出一棵以1为根的树，形式是从节点2开始给出每个节点的父亲节点；

然后是m次操作，操作分为两种，1 v， x， k，表示在以v为根的字数上添加，添加的法则是看这个节点与v节点的距离为i的话，加上x-i*k；

2 v查询节点v的值。

 

发现相加的性质，维护两个树状数组

 

给c1 结点代表的区间都加上x + d[u]*k 给第二个树状数组也加上 d[u]*k

 

假设u是v的父节点 当计算v的时候 可以用$ x + d[u]*k - d[v]*k $

 

正是我们要的$x + k imes (d[u] - d[v])$

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 300010;
const int mod = 1000000007;
vector<int>g[maxn];
LL c[2][maxn],val[2];
int n,m,L[maxn],R[maxn],d[maxn],clk;
void update(int i){
    while(i < maxn){
        c[0][i] += val[0];
        c[1][i] += val[1];
        c[0][i] %= mod;
        c[1][i] %= mod;
        i += i&-i;
    }
}
LL query(int i){
    LL sum[2] = {0},dep = d[i];
    i = L[i];
    while(i > 0){
        sum[0] += c[0][i];
        sum[1] += c[1][i];
        sum[0] %= mod;
        sum[1] %= mod;
        i -= i&-i;
    }
    return ((sum[0] - dep*sum[1])%mod + mod)%mod;
}
void dfs(int u,int dep){
    L[u] = ++clk;
    d[u] = dep;
    for(int i = g[u].size()-1; i >= 0; --i)
        dfs(g[u][i],dep+1);
    R[u] = clk;
}
int main(){
    int u,op,x,y,z;
    while(~scanf("%d",&n)){
        for(int i = clk = 0; i <= n; ++i) g[i].clear();
        for(int i = 2; i <= n; ++i){
            scanf("%d",&u);
            g[u].push_back(i);
        }
        dfs(1,0);
        memset(c,0,sizeof c);
        scanf("%d",&m);
        while(m--){
            scanf("%d%d",&op,&x);
            if(op == 1){
                scanf("%d%d",&y,&z);
                val[0] = ((LL)y + (LL)d[x]*z)%mod;
                val[1] = z;
                update(L[x]);
                val[0] = -val[0];
                val[1] = -val[1];
                update(R[x]+1);
            }else printf("%I64d
",query(x));
        }
    }
    return 0;
}