Codeforces Round #263 (Div. 2) D. Appleman and Tree（树形DP）

题目链接

D. Appleman and Tree

time limit per test :2 seconds

memory limit per test: 256 megabytes

input :standard input

output:standard output

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, ..., 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
Output a single integer — the number of ways to split the tree modulo 1000000007 (109 + 7).

Sample test(s)

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

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[u][0] 不能要，如果要的话 u点的部分会出现不含黑点的情况)

#include <stdio.h>
#include <string.h>
#include <iostream>
#define mod 1000000007

using namespace std ;

struct node
{
int u ;
int v ;
int next ;
}p[100010];
int cnt,head[100010],color[100010] ;
long long dp[100010][2] ;

void addedge(int u,int v)
{
p[cnt].u = u ;
p[cnt].v = v ;
p[cnt].next = head[u] ;
head[u] = cnt ++ ;
}
void DFS(int u)
{
dp[u][color[u]] = 1 ;
for(int i = head[u] ; i+1 ; i = p[i].next)
{
int v = p[i].v ;
DFS(v) ;
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 ;
dp[u][0] = ((dp[u][0] * dp[v][0]) % mod + (dp[u][0] * dp[v][1]) % mod) % mod ;
}
}
int main()
{
int n ,a;
while(~scanf("%d",&n))
{
cnt = 0 ;
memset(head,-1,sizeof(head)) ;
memset(dp,0,sizeof(dp)) ;
for(int i = 1 ; i < n ; i++)
{
scanf("%d",&a) ;
addedge(a,i) ;
}
for(int i = 0 ; i < n ; i++)
scanf("%d",&color[i]) ;
DFS(0) ;
printf("%I64d\n",dp[0][1]) ;
}
return 0 ;
}

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