两个二分图匹配合起来就好了。。中间限制流量为1。
#include<iostream>
#include<cstdio>
#include<cstring>
#define inf 1000000007
using namespace std;
int n,f,d,sum,cnt=1,S,T;
int head[405],cur[405],dis[405],q[405];
int next[50005],list[50005],key[50005];
inline int read()
{
int a=0,f=1; char c=getchar();
while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
while (c>='0'&&c<='9') {a=a*10+c-'0'; c=getchar();}
return a*f;
}
inline void insert(int x,int y,int z)
{
next[++cnt]=head[x];
head[x]=cnt;
list[cnt]=y;
key[cnt]=z;
}
inline bool BFS()
{
memset(dis,-1,sizeof(dis));
dis[S]=1; q[1]=S;
int t=0,w=1,x;
while (t<w)
{
x=q[++t];
for (int i=head[x];i;i=next[i])
if (key[i]&&dis[list[i]]==-1)
dis[list[i]]=dis[x]+1,q[++w]=list[i];
}
return dis[T]!=-1;
}
int find(int x,int flow)
{
if (x==T) return flow;
int w,used=0;
for (int i=cur[x];i;i=next[i])
if (key[i]&&dis[list[i]]==dis[x]+1)
{
w=find(list[i],min(key[i],flow-used));
key[i]-=w; key[i^1]+=w; used+=w;
if (key[i]) cur[x]=i;
if (used==flow) return flow;
}
if (!used) dis[x]=-1;
return used;
}
inline int dinic()
{
int tmp=0;
while (BFS())
{
for (int i=S;i<=T;i++) cur[i]=head[i];
tmp+=find(S,inf);
}
return tmp;
}
int main()
{
n=read(); f=read(); d=read(); S=0; T=n+n+f+d+1;
for (int i=1;i<=n;i++)
{
int T1,T2;
T1=read(); T2=read();
while (T1--)
{
int x=read();
insert(x,i+f,1); insert(i+f,x,0);
}
while (T2--)
{
int x=read();
insert(n+f+i,n+n+f+x,1); insert(n+n+f+x,n+f+i,0);
}
}
for (int i=1;i<=f;i++) insert(S,i,1),insert(i,S,0);
for (int i=1;i<=d;i++) insert(n+n+f+i,T,1),insert(T,n+n+f+i,0);
for (int i=1;i<=n;i++) insert(f+i,f+n+i,1),insert(f+n+i,f+i,0);
cout << dinic() << endl;
return 0;
}