为了满足各种要求来做成网络流也是煞费苦心啊！
引用一下CSDN的某张图，这道题巧妙地把牛拆成了两个点来同时满足两个条件。
AC代码：
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<stack>
#include<cmath>
#include<ctime>
#include<queue>
#include<deque>
#include<list>
#include<map>
#define ffor(i, a, b) for(int i = a; i <= b; i++)
#define rfor(i, a, b) for(int i = a; i >= b; i--)
#define mes(a,b) memset(a, b, sizeof(a))
#define cos(x) cos(x*PI/180.0)
#define sin(x) sin(x*PI/180.0)
#define stop system("pause")
#define see(s,x) cout<<(s)<<'='<<(x)<<endl
#define IMAX 0x7fffffff
#define PI 3.141592654
#define INF 0x3f3f3f3f
#define eps 1e-6
#define lowbit(x) (x&(-x))
typedef long long ll;
const ll mod = (ll)1e9 + 7;
ll max(ll a, ll b) { return(a > b) ? a : b; }
ll min(ll a, ll b) { return(a < b) ? a : b; }
using namespace std;
struct unit
{
int num;
int w;
int rev;
unit() {};
unit(int n, int ww,int r) :num(n), w(ww),rev(r) {};
};
vector<unit> G[501];
int dep[501];
int cur[501];
int n, m, cnt, st, ed;
void add(int x, int y, int z)
{
G[x].push_back(unit(y, z, G[y].size()));
G[y].push_back(unit(x, 0, G[x].size() - 1));
}
bool bfs()
{
mes(dep, 0);
queue<int> q;
q.push(st);
dep[st] = 1;
while (!q.empty())
{
int f = q.front();
q.pop();
for (vector<unit>::iterator it=G[f].begin();it!=G[f].end();it++)
{
unit u = *it;
if (u.w && !dep[u.num])
{
dep[u.num] = dep[f] + 1;
q.push(u.num);
if (u.num == ed)return true;
}
}
}
return false;
}
int dfs(int x, int f)
{
if (x == ed)
return f;
for (int& i = cur[x]; i <= G[x].size() - 1; i++)
{
unit& u = G[x][i];
if (u.w && dep[u.num] == dep[x] + 1)
{
int flow = dfs(u.num, min(f, u.w));
if (flow > 0)
{
u.w -= flow;
G[u.num][u.rev].w += flow;
return flow;
}
}
}
return 0;
}
int dinic()
{
int res = 0;
while (bfs())
{
mes(cur, 0);
res += dfs(st, IMAX);
}
return res;
}
int main()
{
int n, a, b;
cin >> n >> a >> b;
st = 0;
ed = n * 2 + a + b + 1;
ffor(i, n * 2 + 1, n * 2 + a)
add(st, i, 1);
ffor(i, n * 2 + a + 1, n * 2 + a + b)
add(i, ed, 1);
ffor(i, 1, n)
{
add(i, i + n, 1);
int x, y;
scanf("%d%d", &x, &y);
ffor(v, 1, x)
{
int xx;
scanf("%d", &xx);
xx += n * 2;
add(xx, i, 1);
}
ffor(v, 1, y)
{
int yy;
scanf("%d", &yy);
yy += n * 2 + a;
add(i + n, yy, 1);
}
}
printf("%d\n", dinic());
}