codeforces 462 d Appleman and Tree(树形dp)

Appleman and Tree Time Limit:2000MS    Memory Limit:262144KB    64bit IO Format:%I64d & %I64u

Submit Status

Description

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 Input

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<stack>
#include <queue>
#include <set>
#include <map>
using namespace std;
const int inf = 2147483647;
const double PI = acos(-1.0);
const double e = 2.718281828459;
const int mod = 1000000007;
typedef long long LL;
#pragma comment(linker, "/STACK:102400000,102400000")
//freopen("in.txt","r",stdin); //输入重定向,输入数据将从in.txt文件中读取
//freopen("out.txt","w",stdout); //输出重定向,输出数据将保存在out.txt文件中cin
const int N = 100005;
LL dp[N][2];//0 表示没有黑色节点的个数， 1表示已经有1个黑色的个数
int color[N];
vector<int>son[N];
void dp_dfs(int n, int fa)
{
	if (color[n])
		dp[n][1] = 1;
	else
		dp[n][0] = 1;
	for (int i = 0; i < son[n].size(); ++i)
	{
		if (son[n][i] != fa)
		{
			dp_dfs(son[n][i], n);		
			dp[n][1] = ((dp[son[n][i]][0] + dp[son[n][i]][1]) * dp[n][1] % mod + dp[n][0] * dp[son[n][i]][1]) % mod;
			dp[n][0] = (dp[son[n][i]][0] + dp[son[n][i]][1]) * dp[n][0] % mod;
		}
	}
}
int main()
{
	int n, i, j, p;
	while(cin >> n)
	{
		memset(dp, 0, sizeof(dp));
		for (i = 0; i < n; ++i)
			son[i].clear();
		for (i = 1; i < n; ++i)
		{
			scanf("%d", &p);
			son[p].push_back(i);
			son[i].push_back(p);
		}
		for (i = 0; i < n; ++i)
		{
			scanf("%d", &color[i]);
			/*if (color[i] == 0)
				dp[i][1] = 1;*/
		}
		dp_dfs(0, -1);
		printf("%lld\n", dp[0][1]);
	}
}