题解 下落的小球

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 1000010
#define ll long long
#define fir first
#define sec second
#define make make_pair
// #define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
int head[N], size, a[N];
const ll mod=1e9+7;
struct edge{int to, next;}e[N<<1];
inline void add(int s, int t) {e[++size].to=t; e[size].next=head[s]; head[s]=size;}

namespace force{
	int fa[N], tot, ans, p[N];
	bool s[N];
	void solve() {
		for (int i=2; i<=n; ++i) fa[i]=read();
		for (int i=1; i<=n; ++i) a[i]=read();
		for (int i=1; i<=n; ++i) if (a[i]) 
			for (int j=1; j<=a[i]; ++j) p[++tot]=i;
		do {
			for (int i=1; i<=n; ++i) s[i]=1;
			for (int i=1; i<=n; ++i) {
				int now=p[i];
				if (!s[now]) goto jump;
				while (s[fa[now]]) now=fa[now];
				s[now]=0;
			}
			++ans;
			jump: ;
		} while (next_permutation(p+1, p+n+1));
		printf("%d\n", ans);
		exit(0);
	}
}

namespace task{
	int r[N], b[N], s[N];
	ll dp1[N], dp2[N], fac[N], inv[N];
	inline ll C(int n, int k) {return fac[n]*inv[k]%mod*inv[n-k]%mod;}
	void dfs1(int u) {
		s[u]=1; b[u]=a[u];
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			dfs1(v);
			s[u]+=s[v]; b[u]+=b[v];
		}
		r[u]=b[u]-s[u];
	}
	void dfs2(int u) {
		dp1[u]=dp2[u]=1;
		int suma=0, sumb=0;
		for (int i=head[u],v; ~i; i=e[i].next) {
			v = e[i].to;
			dfs2(v);
			dp1[u]=dp1[u]*dp1[v]%mod*C(suma+r[v], r[v])%mod;
			dp2[u]=dp2[u]*dp2[v]%mod*C(sumb+s[v], s[v])%mod;
			suma+=r[v]; sumb+=s[v];
		}
	}
	void solve() {
		memset(head, -1, sizeof(head));
		for (int i=2; i<=n; ++i) add(read(), i);
		for (int i=1; i<=n; ++i) a[i]=read();
		fac[0]=fac[1]=1; inv[0]=inv[1]=1;
		for (int i=2; i<=n; ++i) fac[i]=fac[i-1]*i%mod;
		for (int i=2; i<=n; ++i) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
		for (int i=2; i<=n; ++i) inv[i]=inv[i-1]*inv[i]%mod;
		dfs1(1); dfs2(1);
		// cout<<"dp: "<<dp1[1]<<' '<<dp2[1]<<endl;
		printf("%lld\n", (dp1[1]*dp2[1])%mod);
		exit(0);
	}
}

signed main()
{
	freopen("ball.in", "r", stdin);
	freopen("ball.out", "w", stdout);

	n=read();
	// force::solve();
	task::solve();
	
	return 0;
}