You are given a rooted tree with n nodes. The nodes are numbered 1..n. The root is node 1, and m of the nodes are colored red, the rest are black.
You would like to choose a subset of nodes such that there is no node in your subset which is an ancestor of any other node in your subset. For example, if A is the parent of B and B is the parent of C, then you could have at most one of A, B or C in your subset. In addition, you would like exactly k of your chosen nodes to be red.
If exactly mm of the nodes are red, then for all k = 0..m, figure out how many ways you can choose subsets with k red nodes, and no node is an ancestor of any other node.
Input Format
Each input will consist of a single test case.
Note that your program may be run multiple times on different inputs.
Each test case will begin with a line with two integers n(1≤n≤2×10^5) and m(0≤m≤min(10^3,n)), where n is the number of nodes in the tree, and m is the number of nodes which are red. The nodes are numbered 1..n.
Each of the next n - 1 lines will contain a single integer p(1≤p≤n), which is the number of the parent of this node. The nodes are listed in order, starting with node 2, then node 3, and so on. Node 1 is skipped, since it is the root. It is guaranteed that the nodes form a single tree, with a single root at node 1 and no cycles.
Each of the next m lines will contain single integer r(1≤r≤n). These are the numbers of the red nodes. No value of r will be repeated.
Output Format
Output m + 1 lines, corresponding to the number of subsets satisfying the given criteria with a number of red nodes equal to k = 0..m, in that order. Output this number modulo 10^9 + 7.
样例输入1
4 1 1 1 1 3
样例输出1
5 4
样例输入2
4 4 1 1 1 1 2 3 4
样例输出2
1 4 3 1 0
样例输入3
14 4 1 2 1 2 3 4 5 5 13 8 10 4 4 8 3 12 13
样例输出3
100 169 90 16 0
题意
一棵树，n个点，其中有m个红色，其余为黑点，1为根，选1个点集合，集合内的任意两点不互为祖先，问集合内红色节点的个数为0-m的方案数。
题解
dp[root][m]代表根为root的子树红色节点为m的方案数。
不选root，dp[root][0]=1，考虑子树son的情况，考虑h[M]代表组成M的方案数，相当于每次一颗子树把里面的所有红色节点一个一个压进去，类似于背包。
选root，dp[root][red[root]]++，相当于下面都不能选。
代码
1 #include<bits/stdc++.h>
2 using namespace std;
3 const int N=200005;
4 const int M=1005;
5 const int MD=1000000007;
6 int dp[N][M],h[M],red[N],sz[N];
7 int n,m;
8 vector<int>G[N];
9 void dfs(int u){
10 dp[u][0]=1;
11 sz[u]=red[u];
12 for(int i=0;i<G[u].size();i++) {
13 int v=G[u][i];
14 dfs(v);
15 int up=min(sz[u]+sz[v],m);
16 for(int j=0;j<=up;j++)h[j]=0;
17 for(int j=0;j<=sz[v];j++)
18 for(int k=0;k<=sz[u]&&j+k<=up;k++)
19 h[j+k]=(h[j+k]+1LL*dp[v][j]*dp[u][k])%MD;
20 sz[u]+=sz[v];
21 for(int j=0;j<=sz[u];j++)dp[u][j]=h[j];
22 }
23 dp[u][red[u]]=(dp[u][red[u]]+1)%MD;
24 }
25 int main() {
26 scanf("%d%d",&n,&m);
27 for(int i=2,u;i<=n;i++) {
28 scanf("%d",&u);
29 G[u].push_back(i);
30 }
31 for(int i=0,x;i<m;i++) {
32 scanf("%d",&x);
33 red[x]=1;
34 }
35 dfs(1);
36 for(int i=0;i<=m;i++)
37 printf("%d\n",dp[1][i]);
38 return 0;
39 }
原文地址：https://www.cnblogs.com/taozi1115402474/p/11300110.html