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=0;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[1].beChoosedCount=1;
    num[1].beAbandonedCount=1;
    for (int i = 2; i <= n; ++i)
    {
        
        cin>>num[i].parent;
        num[num[i].parent].son.push_back(i);
        num[num[i].parent].son_count++;
        num[i].beAbandonedCount=1;
        num[i].beChoosedCount=1;
    }

    dfs(1);
    cout<<(num[1].beAbandonedCount+num[1].beChoosedCount-1)%MOD;
    return 0;
}