Codeforces 461B. Appleman and Tree

树形DP。。。

B. Appleman and Tree

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Appleman has a tree with n vertices. Some of the vertices (at least one) are colored black and other vertices are colored white.

Consider a set consisting of k (0 ≤ k < n) edges of Appleman's tree. If Appleman deletes these edges from the tree, then it will split into (k + 1)parts. Note, that each part will be a tree with colored vertices.

Now Appleman wonders, what is the number of sets splitting the tree in such a way that each resulting part will have exactly one black vertex? Find this number modulo 1000000007 (109 + 7).

Input

The first line contains an integer n (2  ≤ n ≤ 105) — the number of tree vertices.

The second line contains the description of the tree: n - 1 integers p0, p1, ..., pn - 2 (0 ≤ pi ≤ i). Where pi means that there is an edge connecting vertex (i + 1) of the tree and vertex pi. Consider tree vertices are numbered from 0 to n - 1.

The third line contains the description of the colors of the vertices: n integers x0, x1, ..., xn - 1 (xi is either 0 or 1). If xi is equal to 1, vertex i is colored black. Otherwise, vertex i is colored white.

Output

Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Sample test(s)

input

3
0 0
0 1 1

output

2

input

6
0 1 1 0 4
1 1 0 0 1 0

output

1

input

10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1

output

27

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
using namespace std;
#define prt(k) cout<<#k"="<<k<<endl;
#define ll long long
const int N=100300;
vector<int> g[N];
const int mod=1e9+7;
ll dp[N][2];
int x[N];
int n;
void dfs(int u)
{
    dp[u][0]=1;
    dp[u][1]=0;
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i];
        dfs(v);
        dp[u][1]=(dp[u][1]*dp[v][0]+dp[u][0]*dp[v][1])%mod;
        dp[u][0]=dp[u][0]*dp[v][0]%mod;
    }
    if(x[u]==1) dp[u][1]=dp[u][0];
    else dp[u][0]=(dp[u][0]+dp[u][1])%mod;
}

int main()
{
    cin>>n;
    for(int i=1;i<n;i++)
    {
        int p; scanf("%d",&p);
        g[p].push_back(i);
    }
    for(int i=0;i<n;i++) scanf("%d",x+i);
    dfs(0);
    ll ans=dp[0][1];
    cout<<ans<<endl;
}