首页 > 代码库 > Codeforces Round #263 (Div. 2) D. Appleman and Tree（树形DP）

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

2024-07-19 20:31:01   222人阅读

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[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 }
View Code

 

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


联系
我们