Codeforces Round #263 (Div. 2) D. Appleman and Tree（树形DP）

浏览数：24 / 时间：2015年06月11日

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, ..., p_n - 2 (0 ≤ p_i ≤ i). Where p_i means that there is an edge connecting vertex (i + 1) of the tree and vertex p_i. 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, ..., x_n - 1 (x_i is either 0 or 1). If x_i 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

input

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

output

input

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

output

题意：对每个节点染色，白或者黑，问你断开某些边，使得每个联通块都恰好只有一个节点时黑色，问有多少种断边方式。

思路：树形DP， dp[i][0]代表到 i 这个点它所在的子树只有一个黑点的情况，dp[i][0] 包含i节点的这部分没有黑点的情况数。

对于每个节点 i，计算到它的一个子树(根节点u) (设连接的边为edge)的时候，dp[i][0] 为dp[i][0] * dp[u][1] + dp[i][0] * dp[u][0], 已处理完的一定要取dp[i][0], 如果取edge 则子树取dp[u][0],如果不取edge, 则子树取dp[u][1].

dp[i][1] 为 dp[i][1] *(dp[u][0] + dp[u][1]) + dp[i][0] *dp[u][1] , 如果处理完的取dp[i][1],edge取的话为dp[u][0], 不取的话为dp[u][1]; 如果处理完的取dp[i][0], edge一定要取且要乘以dp[u][1] (ps: dp[vu][0] 不能要，如果要的话 u点的部分会出现不含黑点的情况)

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #define mod 1000000007
 5 
 6 using namespace std ;
 7 
 8 struct node
 9 {
10     int u ;
11     int v ;
12     int next ;
13 }p[100010];
14 int cnt,head[100010],color[100010] ;
15 long long dp[100010][2] ;
16 
17 void addedge(int u,int v)
18 {
19     p[cnt].u = u ;
20     p[cnt].v = v ;
21     p[cnt].next = head[u] ;
22     head[u] = cnt ++ ;
23 }
24 void DFS(int u)
25 {
26     dp[u][color[u]] = 1 ;
27     for(int i = head[u] ; i+1 ; i = p[i].next)
28     {
29         int v = p[i].v ;
30         DFS(v) ;
31         dp[u][1] = ((dp[u][1] * dp[v][0]) % mod + (dp[u][1] * dp[v][1]) % mod + (dp[u][0] * dp[v][1]) % mod) % mod ;
32         dp[u][0] = ((dp[u][0] * dp[v][0]) % mod + (dp[u][0] * dp[v][1]) % mod) % mod ;
33     }
34 }
35 int main()
36 {
37     int n ,a;
38     while(~scanf("%d",&n))
39     {
40         cnt = 0 ;
41         memset(head,-1,sizeof(head)) ;
42         memset(dp,0,sizeof(dp)) ;
43         for(int i = 1 ; i < n ; i++)
44         {
45             scanf("%d",&a) ;
46             addedge(a,i) ;
47         }
48         for(int i = 0 ; i < n ; i++)
49             scanf("%d",&color[i]) ;
50         DFS(0) ;
51         printf("%I64d\n",dp[0][1]) ;
52     }
53     return 0 ;
54 }