Codeforces 461 B. 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 to1, 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

30 00 1 1

output

2

input

60 1 1 0 41 1 0 0 1 0

output

1

input

100 1 2 1 4 4 4 0 80 0 0 1 0 1 1 0 0 1

output

27

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=100010;const long long int mod=1000000007LL;int n;struct Edge{int to,next;}edge[3*maxn];int Adj[maxn],Size;void init(){Size=0;memset(Adj,-1,sizeof(Adj));}void Add_Edge(int u,int v){edge[Size].to=v;edge[Size].next=Adj[u];Adj[u]=Size++;}int c[maxn];long long int dp[maxn][2];void dfs(int u,int fa){dp[u][c[u]]=1;for(int i=Adj[u];~i;i=edge[i].next){int v=edge[i].to;if(v==fa) continue;dfs(v,u);dp[u][1]=(dp[u][1]*dp[v][1]%mod+dp[u][0]*dp[v][1]%mod+dp[u][1]*dp[v][0]%mod)%mod;dp[u][0]=(dp[u][0]*dp[v][0]%mod+dp[u][0]*dp[v][1]%mod)%mod;}}int main(){init();scanf("%d",&n);for(int i=1;i<n;i++){int v;scanf("%d",&v);Add_Edge(i,v);Add_Edge(v,i);}for(int i=0;i<n;i++)scanf("%d",c+i);dfs(0,0);printf("%I64d\n",dp[0][1]);return 0;}