[网络流24题] 太空飞行计划 - 最小割

（我真是越学越回去了，打回胎重造吧）
经典的最大权闭合子图模型，关键是怎么输出方案。用ISAP的话很显然需要额外搜一遍找出所有与S联通的点就是答案。
笔者第一眼居然以为判下流量就行 傻逼实锤

#include <bits/stdc++.h>
using namespace std;


const int N = 16384, MAXN = 262144;
#define reset(x) memset(x,0,sizeof x)
struct graph
{
    int n,m,M,S,T,head[N],cur[N],dep[N],gap[N],q[N],vis[N];
    long long ans;
    struct ed
    {
        int to,nxt,val,src;
    } edge[MAXN];
    void init(int n0,int m0,int S0,int T0)
    {
        n=n0,m=m0,S=S0,T=T0,M=1,reset(gap);
        reset(head),reset(cur),reset(dep),reset(q);
    }
    int _make(int u,int v,int w)
    {
        edge[++M]=(ed) {v,head[u],w,w},head[u]=M;
        return M;
    }
    int make(int u,int v,int w)
    {
        _make(v,u,0);
        return _make(u,v,w);
    }
    int dfs(int u,int mx)
    {
        if (u==T)
            return mx;
        int num=0,f;
        for (int &i=cur[u],v; i; i=edge[i].nxt)
            if (dep[v=edge[i].to]==dep[u]-1 && (f=edge[i].val))
                if (edge[i].val-=(f=dfs(v,min(mx-num,f))), edge[i^1].val+=f, (num+=f)==mx)
                    return num;
        if (!--gap[dep[u]++])
            dep[S]=n+1;
        return ++gap[dep[u]],cur[u]=head[u],num;
    }
    void _dfs(int p)
    {
        vis[p]=1;
        if(p==T)
            return;
        for(int i=head[p]; i; i=edge[i].nxt)
        {
            int q=edge[i].to;
            if(!vis[q] && edge[i].val)
                _dfs(q);
        }
    }
    void solve()
    {
        for (int i=1; i<=n; ++i)
            cur[i]=head[i];
        ans=0;
        for (gap[0]=n; dep[S]<=n; ans+=dfs(S,0x7fffffff));
    }
} g;

int n,m,t1,t2,t3,t4,tot,lab[100005],mac[100005];
char str[100005];

int main()
{
    cin>>n>>m;
    cin.getline(str,1e+5);
    g.init(n+m+2,0,n+m+1,n+m+2);
    for(int i=1; i<=n; i++)
    {
        cin.getline(str,1e+5);
        stringstream ss(str);
        ss>>t1;
        tot+=t1;
        lab[i] = g.make(n+m+1,i,t1);
        while(ss>>t1)
        {
            g.make(i,n+t1,1e+9);
        }
    }
    for(int i=1; i<=m; i++)
    {
        cin>>t1;
        mac[i] = g.make(n+i,n+m+2,t1);
    }
    g.solve();
    g._dfs(n+m+1);
    int flag = 0;
    for(int i=1; i<=n; i++)
    {
        if(g.vis[i])
        {
            if(flag)
                cout<<" ";
            flag=1;
            cout<<i;
        }
    }
    cout<<endl;
    flag = 0;
    for(int i=1; i<=m; i++)
    {
        if(g.vis[i+n])
        {
            if(flag)
                cout<<" ";
            flag=1;
            cout<<i;
        }
    }
    cout<<endl;
    cout<<tot-g.ans<<endl;
}