Strategic game(树上最小点集覆盖，树形DP)

不取父亲结点，那么必须取儿子节点，这样才能保证父亲和儿子的连边会被覆盖；
取父亲结点，那么儿子节点可取 可不取；
f [ u ] [ 0 ] += f [ v ] [ 1 ] ;
f [ u ] [ 1 ] += min( f [ v ] [ 1 ] , f [ v ] [ 0 ] ) ;

#include<cctype>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=1e5+5;
const int inf=0x3f3f3f3f;
const int mod=1e7+7;
const LL maxn=1e18;
#define fi first
#define se second
#define ls (i<<1)
#define rs ((i<<1)|1)
LL read()
{
    LL x=0,t=1;
    char ch=getchar();
    while(!isdigit(ch)){ if(ch=='-')t=-1; ch=getchar(); }
    while(isdigit(ch)){ x=10*x+ch-'0'; ch=getchar(); }
    return x*t;
}
struct edge
{
    int from,to,next;
    edge(){}
    edge(int ff,int tt,int nn)
    {
        from=ff; to=tt; next=nn;
    }
};
edge e[N<<1];
int f[N][2],head[N],tot;
void init()
{
    tot=0;
    memset(head,0,sizeof(head) );
}
void add(int from,int to)
{
    e[++tot]=edge(from,to,head[from] );
    head[from]=tot;
}
void dfs(int u,int pre)
{
    f[u][0]=0;
    f[u][1]=1;
    for(int i=head[u];i;i=e[i].next)
    {
        int v=e[i].to;
        if(v==pre) continue;
        dfs(v,u);
        f[u][0]+=f[v][1];
        f[u][1]+=min(f[v][1],f[v][0]);
    }
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        init();
        for(int i=1;i<=n;i++)
        {
            int x,y;
            scanf("%d:(%d)",&x,&y);
            //printf("x=%d y=%d
",x,y);
            while(y--)
            {
                int z=read();
                add(z,x); add(x,z);
            }
        }
        dfs(0,-1);
        printf("%d
",min(f[0][0],f[0][1] ) );

    }
    return 0;
}