Codeforces Round #263 (Div. 2) D 树形dp
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D. Appleman and Tree
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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).
Examples
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：
给出根节点为0的树，每个节点被涂成黑色或者白色
现在你需要切k下(1<=k<=n-1)，问有多少种切法使得这k块联通块每块都只有一个节点被涂成黑色
思路：
设置dp[i][2]状态 dp[i][0]表示当前以i为根的子树没有子节点为黑色的种类
dp[i][1]表示当前以i为根的子树存在子节点有黑色的种类(这里指的是已经分好了块的种类)
每次对于一个叶子节点v 有:
dp[i][1]=dp[v][0]*dp[i][1]+dp[v][1]*dp[i][0] 儿子v节点中白色种类*子树当前黑色种类+儿子v黑色种类*子树当前白色种类
dp[i][0]=dp[v][0]*dp[i][0] 儿子节点v中的白色种类*子树当前白色种类
对于每一个子树的遍历完 有：
如果当前子树的根节点为黑色 dp[i][1]=dp[i][0] 满足一个联通块只有一个黑色节点的条件： 当前子树根节点为黑色,所 有黑色的儿子都不能与其相连,种类数为儿子的白色种类数
如果当前子树的根节点为白色 dp[i][0]=dp[i][0]+dp[i][1] 满足所有节点没有黑色的条件：当前的种类+砍掉所有的黑色儿子的种类
代码：
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