Cows are such finicky eaters. Each cow has a preference for certain
foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot
to check his menu against their preferences. Although he might not
be able to stuff everybody, he wants to give a complete meal of
both food and drink to as many cows as possible.

Farmer John has cooked F (1 <= F <= 100) types of foods and prepared
D (1 <= D <= 100) types of drinks. Each of his N (1 <= N <= 100)
cows has decided whether she is willing to eat a particular food
or drink a particular drink. Farmer John must assign a food type
and a drink type to each cow to maximize the number of cows who get
both.

Each dish or drink can only be consumed by one cow (i.e., once food
type 2 is assigned to a cow, no other cow can be assigned food type
2).

输入格式

• Line 1: Three space-separated integers: N, F, and D

• Lines 2..N+1: Each line i starts with a two integers F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i integers following that denote the drinks that cow i will drink.

输出格式

• Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

样例输入

4 3 3
2 2 1 2 3 1
2 2 2 3 1 2
2 2 1 3 1 2
2 1 1 3 3

样例输出

3

提示

INPUT DETAILS:

Cow 1: foods from {1,2}, drinks from {1,3}
Cow 2: foods from {2,3}, drinks from {1,2}
Cow 3: foods from {1,3}, drinks from {1,2}
Cow 4: foods from {1,3}, drinks from {3}

OUTPUT DETAILS:

One way to satisfy three cows is:
Cow 1: no meal
Cow 2: Food #2, Drink #2
Cow 3: Food #1, Drink #1
Cow 4: Food #3, Drink #3
The pigeon-hole principle tells us we can do no better since there are only
three kinds of food or drink. Other test data sets are more challenging, of
course.

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<map>
using std::min;
using std::sort;
using std::pair;
using std::swap;
using std::queue;
using std::vector;
using std::multimap;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) __typeof((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 1; i <= (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int Max_N = 1100;
const int INF = 0x3f3f3f3f;
struct Ford_Fulkerson {
    struct edge { int to, cap, next, rev; }G[Max_N << 2];
    int  N, F, D, tot, head[Max_N];
    bool vis[Max_N], Food[Max_N][Max_N], Drink[Max_N][Max_N];
    inline void init(int n, int f, int d) {
        tot = 0;
        this->N = n, this->F = f, this->D =d;
        cls(head, -1), cls(Food, false), cls(Drink, false);
    }
    inline void add_edge(int u, int v, int cap = 1) {
        G[tot] = (edge){ v, cap, head[u], tot + 1 }; head[u] = tot++;
        G[tot] = (edge){ u,   0, head[v], tot - 1 }; head[v] = tot++;
    }
    inline void built() {
        int x, y, v;
        rep(i, N) {
            scanf("%d %d", &x, &y);
            while(x--) {
                scanf("%d", &v);
                Food[i][v] = true;
            }
            while(y--) {
                scanf("%d", &v);
                Drink[i][v] = true;
            }
        }
    }
    inline int dfs(int u, int t, int f) {
        if(u == t) return f;
        vis[u] = true;
        for(int i = head[u]; ~i; i = G[i].next) {
            edge &e = G[i];
            if(e.cap > 0 && !vis[e.to]) {
                int d = dfs(e.to, t, min(e.cap, f));
                if(d > 0) {
                    e.cap -= d;
                    G[e.rev].cap +=d;
                    return d;
                }
            }
        }
        return 0;
    }
    inline void max_flow(int s, int t) {
        int flow = 0;
        while(true) {
            cls(vis, false);
            int f = dfs(s, t, INF);
            if(!f) break;
            flow += f;
        }
        printf("%d\n", flow);
    }
    inline void solve() {
        int s = 1, t = s + F + 2 * N + D + 1;
        rep(i, F) {
            add_edge(s, s + i);
        }
        rep(i, D) {
            add_edge(s + F + 2 * N + i, t);
        }
        rep(i, N) {
            add_edge(s + F + i, s + F + N + i, 1);
            rep(j, F) {
                if(Food[i][j]) add_edge(s + j, s + F + i);
            }
            rep(j, D) {
                if(Drink[i][j]) add_edge(s + F + N + i, s + F + 2 * N + j);
            }
        }
        max_flow(s, t);
    }
}go;
int main() {
    int n, f, d;
    while(~scanf("%d %d %d", &n, &f, &d)) {
        go.init(n ,f ,d);
        go.built();
        go.solve();
    }
    return 0;
}