摘要：
农夫准备了F（1≤ F≤ 100）食品和D（1≤ D≤ 为他的N（1）杯饮料≤ N≤ 100）头牛。每头牛都有自己喜欢的食物和饮料，每种食物或饮料只能分配给一头牛。有多少头牛能同时得到他们最喜欢的食物和饮料？分析：这是一个经典的网络流量问题。建立一个超级源点，连接到每种食物，建立一个超汇点，连接每种饮料，然后将每头牛分成两个点，一个带食物，一个含饮料，最后运行一次最大流量
题意：农夫为他的 N (1 ≤ N ≤ 100) 牛准备了 F (1 ≤ F ≤ 100)种食物和 D (1 ≤ D ≤ 100) 种饮料。每头牛都有各自喜欢的食物和饮料，
而每种食物或饮料只能分配给一头牛。最多能有多少头牛可以同时得到喜欢的食物和饮料？
析：是一个经典网络流的题，建立一个超级源点，连向每种食物，建立一个超级汇点，连向每种饮料，然后把每头牛拆成两个点，
一个和食物连，一个和饮料连，最后跑一遍最大流即可。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 400 + 5;
const int mod = 1e9;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
struct Edge{
int from, to, cap, flow;
};
struct Dinic{
int n, m, s, t;
vector<Edge> edges;
vector<int> G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
void init(){
edges.clear();
for(int i = 0; i < maxn; ++i) G[i].clear();
}
void addEdge(int from, int to, int cap){
edges.push_back((Edge){from, to, cap, 0});
edges.push_back((Edge){to, from, 0, 0});
m = edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool bfs(){
memset(vis, 0, sizeof vis);
queue<int> q;
q.push(s);
d[s] = 0;
vis[s] = 1;
while(!q.empty()){
int x = q.front(); q.pop();
for(int i = 0; i < G[x].size(); ++i){
Edge &e = edges[G[x][i]];
if(!vis[e.to] && e.cap > e.flow){
vis[e.to] = 1;
d[e.to] = d[x] + 1;
q.push(e.to);
}
}
}
return vis[t];
}
int dfs(int x, int a){
if(x == t || a == 0) return a;
int flow = 0, f;
for(int &i = cur[x]; i < G[x].size(); ++i){
Edge &e = edges[G[x][i]];
if(d[x] + 1 == d[e.to] && (f = dfs(e.to, min(a, e.cap-e.flow))) > 0){
e.flow += f;
edges[G[x][i]^1].flow -= f;
flow += f;
a -= f;
if(a == 0) break;
}
}
return flow;
}
int maxFlow(int s, int t){
this->s = s; this->t = t;
int flow = 0;
while(bfs()){
memset(cur, 0, sizeof cur);
flow += dfs(s, INF);
}
return flow;
}
};
Dinic dinic;
int main(){
int k;
while(scanf("%d %d %d", &n, &m, &k) == 3){
dinic.init();
int s = 0, t = 402;
for(int i = 1; i <= m; ++i) dinic.addEdge(s, i+200, 1);
for(int i = 1; i <= k; ++i) dinic.addEdge(i+300, t, 1);
for(int i = 1; i <= n; ++i){
int f, d;
dinic.addEdge(i, i+100, 1);
scanf("%d %d", &f, &d);
for(int j = 0; j < f; ++j){
int x;
scanf("%d", &x);
dinic.addEdge(x+200, i, 1);
}
for(int j = 0; j < d; ++j){
int x;
scanf("%d", &x);
dinic.addEdge(i+100, x+300, 1);
}
}
printf("%d
", dinic.maxFlow(s, t));
}
return 0;
}