CF 461B Appleman and Tree 树形DP

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：
一棵树，n个节点，编号为0~n-1
每一个节点涂有黑色或者白色，1代表黑色，0代表白色
若在树上去掉k条边，就把树分成k+1部分，每一个部分也是一棵树，若每一部分都有且只有一个节点是黑色，
则这是一个合理的操作。
求合理操作的方案数%(1e9+7)

如果把黑色看成节点的值为1，白色看成节点的值为0
一棵树的值=树上所有节点的值之和
则这道题转化为：
把一棵树分成若干个部分，每一部分的值都为1的方案数。

树形DP

dp[i][0] : 以i为根的子树，i所在部分的值为0的方案数%mod
dp[i][1] : 以i为根的子树，i所在部分的值为1的方案数%mod

以root=0进行DFS
则输出：dp[0][1]

代码：

 #include<cstdio>
 #include<cstring>

 using namespace std;

 const int maxn=1e5+;
 const int mod=1e9+;
 #define ll long long

 struct Edge
 {
     int to,next;
 };
 Edge edge[maxn<<];
 int head[maxn];
 int tot=;
 ll dp[maxn][];
 int cost[maxn];

 void init()
 {
     memset(head,-,sizeof head);
     memset(dp,,sizeof dp);
 }

 void addedge(int u,int v)
 {
     edge[tot].to=v;
     edge[tot].next=head[u];
     head[u]=tot++;
 }

 void solve(int );
 void dfs(int ,int );

 int main()
 {
     init();
     int n;
     scanf("%d",&n);
     for(int i=;i<n;i++)
     {
         int p;
         scanf("%d",&p);
         addedge(i,p);
         addedge(p,i);
     }
     for(int i=;i<n;i++)
     {
         scanf("%d",&cost[i]);
     }
     solve(n);
     return ;
 }

 void solve(int n)
 {
     dfs(,-);
     printf("%d\n",(int)dp[][]);
     return ;
 }

 void dfs(int u,int pre)
 {
     if(cost[u])
     {
         dp[u][]=;
         dp[u][]=;
     }
     else
     {
         dp[u][]=;
         dp[u][]=;
     }
     for(int i=head[u];~i;i=edge[i].next)
     {
         int v=edge[i].to;
         if(v==pre)
             continue;
         dfs(v,u);
         if(cost[u])
         {
             dp[u][]*=(dp[v][]+dp[v][]);
             dp[u][]%=mod;
         }
         else
         {
             dp[u][]=dp[u][]*(dp[v][]+dp[v][])+dp[u][]*dp[v][];
             dp[u][]%=mod;
             dp[u][]*=(dp[v][]+dp[v][]);
             dp[u][]%=mod;
         }
     }
 }