FZU 2277 Change(dfs序+树状数组)

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

题意

给你1个以1为根节点的树，每个节点初始值为0，有下面两个操作

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).

题解

v'为v的子节点

a[v']+=x-(deep[v']-deep[v])*k

a[v']+=x+deep[v]*k-deep[v']*k

对于x+deep[v]*k可以直接更新[s[v],e[v]]

对于-deep[v']*k可以维护一个sumk[v']+=k，最后查询的时候*deep[v']

代码

#include<cstdio>
#include<vector>
using namespace std;

const int maxn=3e5+5;
const int mod=1e9+7;

int s[maxn],e[maxn],deep[maxn],sum[maxn],sumk[maxn],tot,n;
vector<int>G[maxn];
void dfs(int u)
{
    s[u]=++tot;
    for(vector<int>::iterator v=G[u].begin();v!=G[u].end();v++)
        deep[*v]=deep[u]+1,
        dfs(*v);
    e[u]=tot;
}
void update(int x,int add)
{
    for(int i=x;i<=n;i+=(i&-i))sum[i]=(sum[i]+add)%mod;
}
void update1(int x,int add)
{
    for(int i=x;i<=n;i+=(i&-i))sumk[i]=(sumk[i]+add)%mod;
}
int query(int x,int y)
{
    int ret=0,ans=0;
    for(int i=x;i;i-=(i&-i))ret=(ret*1LL+sum[i])%mod;
    for(int i=x;i;i-=(i&-i))ans=(ans*1LL+sumk[i])%mod;
    ans=ans*1LL*deep[y]%mod;
    return (ret+ans)%mod;
}
void init()
{
    tot=0;
    for(int i=1;i<=n;i++)
    {
        sum[i]=sumk[i]=0;
        G[i].clear();
    }
}
int main()
{
    int _,u,v,x,op,Q,k;
    scanf("%d",&_);
    while(_--)
    {
        init();
        scanf("%d",&n);
        for(int i=2;i<=n;i++)
        {
            scanf("%d",&u);
            G[u].push_back(i);
        }
        dfs(1);
        scanf("%d",&Q);
        for(int i=0;i<Q;i++)
        {
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%d%d",&v,&x,&k);
                int ret=(x+deep[v]*1LL*k)%mod;
                update(s[v],ret);
                update(e[v]+1,-ret+mod);
                update1(s[v],-k+mod);
                update1(e[v]+1,k);
            }
            else
            {
                scanf("%d",&v);
                printf("%d
",query(s[v],v));
            }
        }
    }
    return 0;
}