D. Appleman and Tree
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).
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 a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).
3
0 0
0 1 1
2
6
0 1 1 0 4
1 1 0 0 1 0
1
10
0 1 2 1 4 4 4 0 8
0 0 0 1 0 1 1 0 0 1
27
题意：对每个节点染色，白或者黑，问你断开某些边，使得每个联通块都恰好只有一个节点时黑色，问有多少种断边方式。
思路 ：树形DP， dp[i][0]代表到 i 这个点它所在的子树只有一个黑点的情况，dp[i][0] 包含i节点的这部分没有黑点的情况数。
对于每个节点 i，计算到它的一个子树(根节点u) (设连接的边为edge)的时候，dp[i][0] 为dp[i][0] * dp[u][1] + dp[i][0] * dp[u][0], 已处理完的一定要取dp[i][0], 如果取edge 则子树取dp[u][0],如果不取edge, 则子树取dp[u][1].
dp[i][1] 为 dp[i][1] *(dp[u][0] + dp[u][1]) + dp[i][0] *dp[u][1] , 如果处理完的取dp[i][1],edge取的话为dp[u][0], 不取的话为dp[u][1]; 如果处理完的取dp[i][0], edge一定要取且要乘以dp[u][1] (ps: dp[vu][0] 不能要，如果要的话 u点的部分会出现不含黑点的情况)
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 #define mod 1000000007 5 6 using namespace std ; 7 8 struct node 9 { 10 int u ; 11 int v ; 12 int next ; 13 }p[100010]; 14 int cnt,head[100010],color[100010] ; 15 long long dp[100010][2] ; 16 17 void addedge(int u,int v) 18 { 19 p[cnt].u = u ; 20 p[cnt].v = v ; 21 p[cnt].next = head[u] ; 22 head[u] = cnt ++ ; 23 } 24 void DFS(int u) 25 { 26 dp[u][color[u]] = 1 ; 27 for(int i = head[u] ; i+1 ; i = p[i].next) 28 { 29 int v = p[i].v ; 30 DFS(v) ; 31 dp[u][1] = ((dp[u][1] * dp[v][0]) % mod + (dp[u][1] * dp[v][1]) % mod + (dp[u][0] * dp[v][1]) % mod) % mod ; 32 dp[u][0] = ((dp[u][0] * dp[v][0]) % mod + (dp[u][0] * dp[v][1]) % mod) % mod ; 33 } 34 } 35 int main() 36 { 37 int n ,a; 38 while(~scanf("%d",&n)) 39 { 40 cnt = 0 ; 41 memset(head,-1,sizeof(head)) ; 42 memset(dp,0,sizeof(dp)) ; 43 for(int i = 1 ; i < n ; i++) 44 { 45 scanf("%d",&a) ; 46 addedge(a,i) ; 47 } 48 for(int i = 0 ; i < n ; i++) 49 scanf("%d",&color[i]) ; 50 DFS(0) ; 51 printf("%I64d\n",dp[0][1]) ; 52 } 53 return 0 ; 54 }