[题目链接]

思路：最大流变形—顶点也有容量限制的情况！

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
const int inf=0x3f3f3f3f;
const int Max_v=1e5+10;

int n,f,d;
struct edge{
    int to,cap,rev;
    edge(int t,int c,int r):to(t),cap(c),rev(r){}
};
vector<edge>G[Max_v];
int level[Max_v];
int iter[Max_v];

void add_edge(int from,int to,int cap){
    G[from].push_back(edge(to,cap,G[to].size()));
    G[to].push_back(edge(from,0,G[from].size()-1));
}

void bfs(int s){
    memset(level,-1,sizeof(level));
    queue<int>que;
    level[s]=0;
    que.push(s);
    while(!que.empty()){
        int u=que.front();que.pop();
        for(int i=0;i<G[u].size();i++){
            edge e=G[u][i];
            if(e.cap>0&&level[e.to]<0){
                level[e.to]=level[u]+1;
                que.push(e.to);
            }
        }
    }
}

int dfs(int u,int t,int f){
    if(u==t)return f;
    for(int &i=iter[u];i<G[u].size();i++){
        edge &e=G[u][i];
        if(e.cap>0&&level[u]<level[e.to]){
            int d=dfs(e.to,t,min(f,e.cap));
            if(d>0){
                e.cap-=d;
                G[e.to][e.rev].cap+=d;
                return d;
            }
        }
    }
    return -1;
}

int max_flow(int s,int t){
    int flow=0;
    while(true){
        bfs(s);
        if(level[t]<0)return flow;
        memset(iter,0,sizeof(iter));
        int dist;
        while((dist=dfs(s,t,inf))>0){
            flow+=dist;
        }
    }
}

int main()
{
    scanf("%d%d%d",&n,&f,&d);
    int s=0,t=550; //源点s汇点t
    for(int i=1;i<=n;i++){ //牛[1,100]牛[101,200]之间连边
        add_edge(i,i+100,1);
    }
    for(int i=1;i<=f;i++){ //源点s(0)和食物[201,300]之间连边
        add_edge(s,i+200,1);
    }
    for(int i=1;i<=d;i++){ //饮料[301,400]和汇点t(550)之间连边
        add_edge(i+300,t,1);
    }
    int a,b,c,d;
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a,&b);
        for(int j=0;j<a;j++){ //食物[201,300]和牛[1,100]之间连边
            scanf("%d",&c);
            add_edge(c+200,i,1);
        }
        for(int j=0;j<b;j++){ //牛[101,200]和饮料[301,400]之间连边
            scanf("%d",&c);
            add_edge(i+100,c+300,1);
        }
    }
    printf("%d\n",max_flow(s,t));
    return 0;
}