G将军的敢死队——树状DP

当前节点的两种情况：

1.beChoosed = {son.beAbandoned乘积}　　//当前节点选中的情况下，子节点都不能选

2.beAbandoned = {(son.beAbandoned + son.beChoosed)乘积} //当前节点不选的情况下，子节点所有情况都算上

ps：最终要减去一种所有节点都未被选中的情况  －1

 #include <iostream>
 #include <stdio.h>
 #include <list>
 #define MOD 10007
 #define MAXN 100002
 using namespace std;
 int n;
 struct node{
     int parent;
     list<int>son;
     int son_count;
     int beChoosedCount,beAbandonedCount;
 }num[MAXN];

 void dfs(int id)
 {
     for(int i=;i<num[id].son_count;++i)
     {
         int tmp = num[id].son.front();
         dfs(tmp);
         num[id].son.pop_front();
         num[id].beChoosedCount *= num[tmp].beAbandonedCount%MOD;
         num[id].beAbandonedCount *= (num[tmp].beChoosedCount+num[tmp].beAbandonedCount)%MOD;
     }

 }
 int main()
 {
     cin>>n;
     num[].beChoosedCount=;
     num[].beAbandonedCount=;
     for (int i = ; i <= n; ++i)
     {

         cin>>num[i].parent;
         num[num[i].parent].son.push_back(i);
         num[num[i].parent].son_count++;
         num[i].beAbandonedCount=;
         num[i].beChoosedCount=;
     }

     dfs();
     cout<<(num[].beAbandonedCount+num[].beChoosedCount-)%MOD;
     return ;
 }