2016 ACM/ICPC ECNA Regional I.Waif Until Dark(最大流)

这是一道ECNA的16年题，问有n个小朋友，m个玩具，不同孩子有不同喜好的玩具，每个玩具可能属于一个类别，同一类别的玩具最多只能用一定次数，问最大匹配

这个就很裸的二分图，掏出dinic板子，首先最后问的是孩子，所以最后一层肯定是孩子，然后有玩具，所以玩具是和孩子连边的，有种类，那种类和玩具连边，然后如果不属于任何一组的玩具边的容量就是1，与源连上，如果属于的话就和组连上，容量为最大限制，孩子和终点连边，跑一次dinic就行了，easy

#include <bits/stdc++.h>
using namespace std;
#define limit (90000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n,m,vs,ve,p;
int layer[limit],head[limit], cnt;
struct node{
    int to ,next;
    ll flow, w;
}edge[limit];
ll max_flow;
void add_one(int u , int v, ll flow = 0){
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    edge[cnt].flow = flow;
    edge[cnt].w = 0;
    head[u] = cnt++;
}
inline void add(int u, int v, ll flow){
    add_one(u,v,flow);
    add_one(v, u,0);
}
inline void init(bool flag = true){
    if(flag){
        memset(head, -1, sizeof(head));
        cnt = 0;
    }else{
        memset(layer, -1, sizeof(layer));
    }
}
inline bool bfs(){
    init(false);
    queue<int>q;
    layer[vs] = 0;//从第0层开始
    q.push(vs);
    while (q.size()){
        int u = q.front();
        q.pop();
        traverse(u){
            int v = edge[i].to,flow = edge[i].flow;
            if(layer[v] == -1 && flow > 0){
                layer[v] = layer[u] + 1;//迭代加深
                q.push(v);
            }
        }
    }
    return ~layer[ve];
}
ll dfs(int u, ll flow){
    if(u == ve)return flow;
    ll rev_flow = 0,min_flow;
    traverse(u){
        int v =edge[i].to;
        ll t_flow = edge[i].flow;
        if(layer[v] == layer[u] + 1 && t_flow > 0){
            min_flow = dfs(v, min(flow, t_flow));
            flow -= min_flow;
            edge[i].flow -= min_flow;
            rev_flow += min_flow;
            edge[i^1].flow += min_flow;
            if(!flow)break;
        }
    }
    if(!rev_flow)layer[u] = -1;
    return rev_flow;
}
void dinic(){
    while (bfs()){
        max_flow += dfs(vs,inf);
    }
}
int vis[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    init();
    n = read(), m = read(), p = read();
    vs = 80001, ve = vs + 1;
    rep(i, 1,n){
        add(m + p + i, ve, 1);//孩子
    }
    rep(i ,1,n){
        //连孩子和玩具
        int op = read();
        while (op--){
            int num_toy = read();
            add(p + num_toy , m + p + i, 1);
        }
    }
    rep(q,1,p){
        int l = read();
        while (l--){
            int x = read();
            vis[x] = 1;
            add(q, p + x, 1);
        }
        int r = read();
        add(vs,q,r);
    }
    rep(i ,1,m)if(!vis[i])add(vs, p + i, 1);
    dinic();
    write(max_flow);
    return 0;
}