[CF543D]Road Improvement

题目大意：给定一个无根树，给每条边黑白染色，求出每个点为根时，其他点到根的路径上至多有一条黑边的染色方案数，模$1e9+7$。

题解：树形$DP$不难想到，记$f_u$为以$1$为根时，以$u$为根的子树的方案数，$f_u=\prod\limits_{v\in son_u}(f_v+1）$

换根也很简单。

但是这题卡模数，换根时要求逆元，其中$f_u$可能等于$1e9+6$，加一后变成$0$，无法求逆。可以求前缀积和后缀积转移

卡点：原$dp$写错

C++ Code：

#include <cstdio>
#include <vector>
#include <cctype>
namespace R {
	int x, ch;
	inline int read() {
		ch = getchar();
		while (isspace(ch)) ch = getchar();
		for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
		return x;
	}
}
using R::read;

#define maxn 200010
const int mod = 1e9 + 7;
int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
}

int n;
int f[maxn], g[maxn], ans[maxn], l[maxn], r[maxn];
void dfs(int u, int fa = 0) {
	f[u] = 1;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa) {
			dfs(v, u);
			f[u] = static_cast<long long> (1 + f[v]) * f[u] % mod;
		}
	}
}
void dfs1(int u, int fa = 0) {
	std::vector<int> S;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa) S.push_back(v);
	}
	if (S.size()) {
		l[*S.begin()] = g[u];
		for (std::vector<int>::iterator it = S.begin() + 1; it != S.end(); it++) {
			l[*it] = static_cast<long long> (l[*(it - 1)]) * (f[*(it - 1)] + 1) % mod;
		}
		r[*(S.end() - 1)] = 1;
		if (S.begin() + 1 != S.end()) {
			for (std::vector<int>::iterator it = S.end() - 2; true; it--) {
				r[*it] = static_cast<long long> (r[*(it + 1)]) * (f[*(it + 1)] + 1) % mod;
				if (it == S.begin()) break;
			}
		}
	}
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa) {
			g[v] = (static_cast<long long> (l[v]) * r[v] + 1) % mod; 
			ans[v] = static_cast<long long> (g[v]) * f[v] % mod;
			dfs1(v, u);
		}
	}
}

int main() {
	n = read();
	for (int i = 1, x; i < n; i++) {
		x = read();
		add(i + 1, x);
		add(x, i + 1);
	}
	dfs(1);
	ans[1] = f[1]; g[1] = 1;
	dfs1(1);
	for (int i = 1; i <= n; i++) {
		printf("%d", ans[i]);
		putchar(i == n ? '\n' : ' ');
	}
	return 0;
}