題意：

有M個豬圈，每個豬圈裡初始時有若干頭豬。一開始所有豬圈都是關閉的。依次來了N個顧客，每個顧客分別會開啟指定的幾個豬圈，從中買若干頭豬。每個顧客分別都有他能夠買的數量的上限。每個顧客走後，他開啟的那些豬圈中的豬，都可以被任意地調換到其它開著的豬圈裡，然後所有豬圈重新關上。問總共最多能賣出多少頭豬。（1 <= N <= 100, 1 <= M <= 1000）

解析：

因為是依次 所以對於當前第i個顧客 所對應的豬圈 向前 i - 1 個顧客種有對應這個豬圈的顧客的其它豬圈 就好啦

#include <iostream>
#include <cstdio>
#include 
 
<sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define
 
 rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define
 
 rf(a) scanf("%f", &a)
#define pd(a) printf("%d\n", a)
#define plld(a) printf("%lld\n", a)
#define pc(a) printf("%c\n", a)
#define ps(a) printf("%s\n", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 1e5 + 10, INF = 0x7fffffff;
int n, m, s, t;
vector<int> g[maxn], f[maxn];
int head[maxn], cur[maxn], vis[maxn], d[maxn], cnt, nex[maxn << 1];

struct node
{
    int u, v, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    nex[cnt] = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int c)
{
    add_(u, v, c);
    add_(v, u, 0);
}

bool bfs()
{
    queue<int> Q;
    mem(d, 0);
    Q.push(s);
    d[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop();
        for(int i = head[u]; i != -1; i = nex[i])
        {
            int v = Node[i].v;
            if(!d[v] && Node[i].c > 0)
            {
                d[v] = d[u] + 1;
                Q.push(v);
                if(v == t) return 1;
            }
        }
    }
    return d[t] != 0;
}

int dfs(int u, int cap)
{
    int ret = 0;
    if(u == t || cap == 0)
        return cap;
    for(int &i = cur[u]; i != -1; i = nex[i])
    {
        int v = Node[i].v;
        if(d[v] == d[u] + 1 && Node[i].c > 0)
        {
            int V = dfs(v, min(cap, Node[i].c));
            Node[i].c -= V;
            Node[i ^ 1].c += V;
            ret += V;
            cap -= V;
            if(cap == 0) break;
        }
    }
    if(cap > 0) d[u] = -1;
    return ret;
}

int Dinic()
{
    int ans = 0;
    while(bfs())
    {
        memcpy(cur, head, sizeof head);
        ans += dfs(s, INF);
    }
    return ans;
}

int main()
{
    while(scanf("%d %d", &m, &n) != EOF)
    {
        mem(head, -1);
        cnt = 0;
        for(int i = 0; i < maxn; i++) g[i].clear(), f[i].clear();
        int w, x, v;
        s = 0, t = n + m + 10;
        rap(i, 1, m)
        {
            rd(w);
            add(i, t, w);
        }
        rap(i, 1, n)
        {
            rd(x);
            rap(j, 1, x)
            {
                rd(v);
                add(m + i, v, INF);
                for(int k = 0; k < g[v].size(); k++)
                {
                    for(int p = 0; p < f[g[v][k]].size(); p++)
                        add(m + i, f[g[v][k]][p], INF);

                }
                g[v].push_back(i);
                f[i].push_back(v);
            }
            rd(w);
            add(s, m + i, w);
        }
        pd(Dinic());


    }

    return 0;
}