題意：

n個英雄，m個怪獸，第i個英雄可以打第i個集合裡的怪獸，一個怪獸可以在多個集合裡

有k瓶藥水，每個英雄最多喝一次，可以多打一隻怪獸，求最多打多少隻

n,m,k<=500

思路：

最大流，建圖方式：

程式碼：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
//#include<stack>
#include<queue>
#include
 
<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>
    
#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define
 
 rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-6;
const int mod = 1e9+7;
const
 
 int maxn = 2e6+100;
const int maxm = 2e6+100;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int head[maxn],d[maxn];//層
int ver[maxm],edge[maxm],Next[maxm];//edge[i]: c for edge_i
int n, m, s, t, tot, maxflow;
queue<int>q;
int vis[maxn];//出現過
void add(int x, int y, int z){
    ver[++tot]=y,edge[tot]=z,Next[tot]=head[x],head[x]=tot;
    
    ver[++tot]=x,edge[tot]=0,Next[tot]=head[y],head[y]=tot;
}

bool bfs(){
    mem(d,0);
    while(!q.empty())q.pop();
    q.push(s);
    d[s]=1;
    while(!q.empty()){
        int x = q.front();
        q.pop();
        for(int i = head[x]; i; i = Next[i]){
            if(edge[i] && !d[ver[i]]){
            q.push(ver[i]);
            d[ver[i]] = d[x] + 1;
            if(ver[i] == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow){
    if(x==t) return flow;
    int rest = flow, k;
    for(int i = head[x]; i; i = Next[i]){
        if(edge[i] && d[ver[i]] == d[x]+1){
            k = dinic(ver[i], min(rest, edge[i]));
            if(!k) d[ver[i]] = 0;
            edge[i] -= k;
            edge[i^1] += k;
            rest -= k;
        }
    }
    return flow - rest;
}

int main(){
    int num;
    scanf("%d %d %d", &n, &m, &num);
    tot = 1;
    //scanf("%d %d", &s, &t);
    s = 1;
    t = n+m+3;
    add(s,2,num);
    for(int i = 1; i <= n; i++){
        int sz, x;
        add(s,2+i,1);
        add(2,2+i,1);
        scanf("%d", &sz);
        for(int j = 1; j <= sz; j++){
            scanf("%d", &x);
            add(2+i,2+n+x,1);
        }
    }
    for(int i = 1; i <= m; i++){
        add(2+n+i,t,1);
    }
    int flow = 0;
    maxflow=0;
    while(bfs()){
        while(1){
            flow = dinic(s,inf);
            if(!flow)break;
            maxflow+=flow;
        }
    }
    printf("%d\n",maxflow);
    return 0;
}
/*
3 5 0
4 1 2 3 5
2 2 5
2 1 2


5 10 2
2 3 10
5 1 3 4 6 10
5 3 4 6 8 9
3 1 9 10
5 1 3 6 7 10

 */