Less Time, More profit 最大权闭合子图（最大流最小割）

The city planners plan to build N plants in the city which has M shops.

Each shop needs products from some plants to make profit of proiproi units.

Building ith plant needs investment of payipayi units and it takes titi days.

Two or more plants can be built simultaneously, so that the time for building multiple plants is maximum of their periods(titi).

You should make a plan to make profit of at least L units in the shortest period. 

InputFirst line contains T, a number of test cases.

For each test case, there are three integers N, M, L described above.

And there are N lines and each line contains two integers payipayi, titi(1<= i <= N).

Last there are M lines and for each line, first integer is proiproi, and there is an integer k and next k integers are index of plants which can produce material to make profit for the shop.

1 <= T <= 30 
1 <= N, M <= 200 
1≤L,ti≤10000000001≤L,ti≤1000000000 
1≤payi,proi≤300001≤payi,proi≤30000 
OutputFor each test case, first line contains a line “Case #x: t p”, x is the number of the case, t is the shortest period and p is maximum profit in t hours. You should minimize t first and then maximize p.

If this plan is impossible, you should print “Case #x: impossible” 
Sample Input

2
 
1 1 2
1 5
3 1 1
 
1 1 3
1 5
3 1 1

Sample Output

Case #1: 5 2
Case #2: impossible

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 405
#define L 31
#define INF 1000000009
#define eps 0.00000001
#define sf(a) scanf("%lld",&a)
struct plant
{
    LL pay, time;
    LL id;
    bool operator<(const plant& rhs) const
    {
        return time < rhs.time;
    }
};
struct shop
{
    LL cnt, time, pro;
    LL pl[MAXN];
};
shop s[MAXN];
plant p[MAXN];
LL g[MAXN << ][MAXN << ];
LL level[MAXN<<];
LL T, n, m, l, st, ed, ans, tmp;
bool bfs()
{
    memset(level, -, sizeof(level));
    level[st] = ;
    queue<LL> q;
    q.push(st);
    while (!q.empty())
    {
        LL f = q.front();
        q.pop();
        for (LL i = ; i <= ed; i++)
        {
            if (level[i] == - && g[f][i] > )
            {
                level[i] = level[f] + ;
                q.push(i);
            }
        }
    }
    return level[ed] > ;
}
LL dinic(LL k, LL low)
{
    if (k == ed)return low;
    LL a;
    for (LL i = ; i <= ed; i++)
    {
        if (level[i] == level[k] +  && g[k][i] >  && (a = dinic(i, min(low, g[k][i]))))
        {
            g[k][i] -= a;
            g[i][k] += a;
            return a;
        }
    }
    return ;
}
void solve()
{
    ans = ;
    while (bfs())
    {
        while (tmp = dinic(st, INF))
            ans += tmp;
    }
}
int main()
{
    sf(T);
    for (LL cas = ; cas <= T; cas++)
    {
        sf(n), sf(m), sf(l);
        for (LL i = ; i <= n; i++)
        {
            sf(p[i].pay), sf(p[i].time);
            p[i].id = i;
        }
        for (LL i = ; i <= m; i++)
        {
            sf(s[i].pro);
            sf(s[i].cnt);
            s[i].time = ;
            for (LL j = ; j < s[i].cnt; j++)
            {
                sf(s[i].pl[j]);
                s[i].time = max(s[i].time, p[s[i].pl[j]].time);
            }
        }
        sort(p + , p +  + n);
        bool f = false;
        st = n + m + , ed = st + ;
        printf("Case #%lld: ", cas);
        for (LL i = ; i <= n; i++)
        {
            memset(g, , sizeof(g));
            for (LL j = ; j <= i; j++)
                g[p[j].id][ed] = p[j].pay;
            LL tot = ;
            for (LL j = ; j <= m; j++)
            {
                if (s[j].time <= p[i].time)
                {
                    tot += s[j].pro;
                    g[st][j + n] = s[j].pro;
                    for (LL k = ; k < s[j].cnt; k++)
                        g[j + n][s[j].pl[k]] = INF;
                }
            }
            solve();
            ans = tot - ans;
            if (ans >= l)
            {
                printf("%lld %lld\n", p[i].time, ans);
                f = true;
                break;
            }
        }
        if (!f)
            printf("impossible\n");
    }
}