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如果只有两个的话就是裸的二分图匹配

然而这里有三个元素

由于食物和饮料互相没有关系，只和牛有关系

那么我们把牛拆点x->y cap = 1

然后饮料和食物分别限制x y就行了

#include<bits/stdc++.h>
using namespace std;
const int N = 410, inf = 1e9;
int rd()
{
    int x = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') f = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}
int n, m, k, source, sink, cnt = 1, tot;
int a[N], d[N], head[N], iter[N], Map[N][N];
struct edge {
    int nxt, to, f;
} e[N * N];
bool bfs()
{
    queue<int> q;
    memset(d, -1, sizeof(d));
    d[source] = 0;
    q.push(source);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        for(int i = head[u]; i; i = e[i].nxt) if(d[e[i].to] == -1 && e[i].f) 
        {
            d[e[i].to] = d[u] + 1;
            q.push(e[i].to);
        }
    }
    return d[sink] != -1;
}
int dfs(int u, int delta)
{
    if(u == sink) return delta;
    int ret = 0;
    for(int &i = iter[u]; i && delta; i = e[i].nxt) if(d[e[i].to] == d[u] + 1 && e[i].f)
    {
        int x = dfs(e[i].to, min(delta, e[i].f));
        ret += x;
        delta -= x;
        e[i].f -= x;
        e[i ^ 1].f += x;    
    }
    return ret;
}
int dinic() 
{
    int ret = 0;
    while(bfs())
    {
        for(int i = source; i <= sink; ++i) iter[i] = head[i];
        ret += dfs(source, inf);
    }
    return ret;
}
void link(int u, int v, int f)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
    e[cnt].f = f;
}
void insert(int u, int v, int f)
{
    link(u, v, f);
    link(v, u, 0);
}
int main()
{
    scanf("%d%d%d", &n, &m, &k);
    sink = 2 * n + m + k + 1;
    for(int i = 1; i <= n; ++i)
    {
        insert(i, i + n, 1);
        int n1, n2;
        scanf("%d%d", &n1, &n2);
        while(n1--)
        {
            int x;
            scanf("%d", &x);
            insert(x + 2 * n, i, 1);
        }
        while(n2--)
        {
            int x;
            scanf("%d", &x);
            insert(i + n, x + 2 * n + m, 1);
        }
    }
    for(int i = 1; i <= m; ++i) insert(source, i + 2 * n, 1);
    for(int i = 1; i <= k; ++i) insert(i + 2 * n + m, sink, 1);
    printf("%d\n", dinic());
    return 0;
}