CCC 2011 Unfriend

题面传送门
显然树形 d p dp dp
转移时直接考虑这颗子树选或不选。 d p v = d p v × ( d p u + 1 ) [ u ∈ v ] dp_v=dp_v\times (dp_u+1)[u∈v] dpv​=dpv​×(dpu​+1)[u∈v]
代码实现:

#include<cstdio>
#define beg int cur=s.h[x]
#define end cur
#define go cur=tmp.z
using namespace std;
int n,m,k,x,y,z,dp[39],in[39];
struct yyy{int to,z;};
struct ljb{
    int head,h[39];
    yyy f[39];
    inline void add(int x,int y){
        f[++head]=(yyy){y,h[x]};
        h[x]=head;
    }
}s;
inline void dfs(int x){
    yyy tmp;dp[x]=1;
    for(beg;end;go){
        tmp=s.f[cur];
        dfs(tmp.to);
        dp[x]*=(dp[tmp.to]+1);
    }
}
int main(){
    register int i;
    scanf("%d",&n);
    for(i=1;i<n;i++) scanf("%d",&x),s.add(x,i),in[i]++;
    for(i=1;i<=n;i++){
        if(!in[i]) dfs(i),printf("%d\n",dp[i]);
    }
}