Codeforces Round #263 (Div. 2)D（树形DP）

D. 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

题意：给了一棵树以及每个节点的颜色，1代表黑，0代表白，要求的是，如果将这棵树拆成k棵树，使得每棵树恰好有一个黑色节点

思路：树形dp

          dp[u][0]表示以u为根的子树对父亲的贡献为0

          dp[u][1]表示以u为根的子树对父亲的贡献为1

          我现在假设u为白色，它的子树有x，y，z，那么有

          dp[u][1]+=dp[x][1]*dp[y][0]*dp[z][0]+dp[x][0]*dp[y][1]*dp[z][0]+dp[x][0]*dp[y][0]*dp[z][1]

          dp[u][0]+=dp[x][0]*dp[y][0]*dp[z][0]

          然后判断u的颜色，假设u的父亲为p

         1.为黑色，不切断边（u，p），那么dp[u][1]=dp[u][0]，切断（u，p），那么dp[u][0]不变

         2.为白色，如果切断（u，p），dp[u][0]还要加上dp[u][1]