Fill a DP table such as the following bottom-up:
- DP[v][0] = the number of ways that the subtree rooted at vertex v has no black vertex.
- DP[v][1] = the number of ways that the subtree rooted at vertex v has one black vertex.
The recursion pseudo code is folloing:
DFS(v):
DP[v][0] = 1
DP[v][1] = 0
foreach u : the children of vertex v
DFS(u)
DP[v][1] *= DP[u][0]
DP[v][1] += DP[v][0]*DP[u][1]
DP[v][0] *= DP[u][0]
if x[v] == 1:
DP[v][1] = DP[v][0]
else:
DP[v][0] += DP[v][1]The answer is DP[root][1]
#include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set> #include <string> #include <queue> #include <stack> #include <bitset> using namespace std; #define pb(x) push_back(x) #define ll long long #define mk(x, y) make_pair(x, y) #define lson l, m, rt<<1 #define mem(a) memset(a, 0, sizeof(a)) #define rson m+1, r, rt<<1|1 #define mem1(a) memset(a, -1, sizeof(a)) #define mem2(a) memset(a, 0x3f, sizeof(a)) #define rep(i, n, a) for(int i = a; i<n; i++) #define fi first #define se second typedef pair<int, int> pll; const double PI = acos(-1.0); const double eps = 1e-8; const int mod = 1e9+7; const int inf = 1061109567; const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; const int maxn = 1e5+2; ll dp[maxn][2]; int head[maxn], num, k, a[maxn]; struct node { int to, nextt; }e[maxn*2]; void add(int u, int v) { e[num].to = v, e[num].nextt = head[u], head[u] = num++; } void init() { num = 0; mem1(head); } void dfs(int u, int fa) { dp[u][1] = 0; dp[u][0] = 1; for(int i = head[u]; ~i; i = e[i].nextt) { int v = e[i].to; if(v == fa) continue; dfs(v, u); dp[u][1] = dp[u][1]*dp[v][0]%mod; dp[u][1] = (dp[u][1]+dp[u][0]*dp[v][1]%mod)%mod; dp[u][0] = dp[u][0]*dp[v][0]%mod; } if(a[u]) { dp[u][1] = dp[u][0]; } else { dp[u][0] = (dp[u][0]+dp[u][1])%mod; } } int main() { int n, x, y; cin>>n; init(); for(int i = 1; i<n; i++) { scanf("%d", &x); add(x, i); add(i, x); } for(int i = 0; i<n; i++) scanf("%d", &a[i]); dfs(0, -1); cout<<dp[0][1]<<endl; return 0; }