The Perfect Stall

题目链接

• 题意：
  

  有n个奶牛和m个谷仓，现在每个奶牛有自己喜欢去的谷仓，并且它们只会去自己喜欢的谷仓吃东西，问最多有多少奶牛能够吃到东西
  输入第一行给出n与m，接着n行，每行第一个数代表这个奶牛喜欢的谷仓的个数P，后面接着P个数代表这个奶牛喜欢哪个谷仓
  N (0 <= N <= 200) and M (0 <= M <= 200)

• 分析：
  明显的二分图，这里练习网络流

struct Edge{    int from, to, cap, flow;    bool operator< (const Edge& rhs) const    {        return from < rhs.from || (from == rhs.from && to < rhs.to);    }};const int MAXV = 410;struct ISAP{    int n, m, s, t;    vector<Edge> edges;    vector<int> G[MAXV];   // 邻接表，G[i][j]表示结点i的第j条边在e数组中的序号    bool vis[MAXV];        // BFS使用    int d[MAXV];           // 从起点到i的距离    int cur[MAXV];        // 当前弧指针    int p[MAXV];          // 可增广路上的上一条弧    int num[MAXV];        // 距离标号计数    void AddEdge(int from, int to, int cap)    {        edges.push_back((Edge) { from, to, cap, 0 });        edges.push_back((Edge) { to, from, 0, 0 });        m = edges.size();        G[from].push_back(m-2);        G[to].push_back(m-1);    }    bool BFS()    {        memset(vis, 0, sizeof(vis));        queue<int> Q;        Q.push(t);        vis[t] = 1;        d[t] = 0;        while(!Q.empty())        {            int x = Q.front();            Q.pop();            REP(i, G[x].size())            {                Edge& e = edges[G[x][i]^1];                if(!vis[e.from] && e.cap > e.flow)                {                    vis[e.from] = 1;                    d[e.from] = d[x] + 1;                    Q.push(e.from);                }            }        }        return vis[s];    }    void ClearAll(int n)    {        this->n = n;        REP(i, n)            G[i].clear();        edges.clear();    }    void ClearFlow()    {        REP(i, edges.size())            edges[i].flow = 0;    }    int Augment()    {        int x = t, a = INF;        while(x != s)        {            Edge& e = edges[p[x]];            a = min(a, e.cap-e.flow);            x = edges[p[x]].from;        }        x = t;        while(x != s)        {            edges[p[x]].flow += a;            edges[p[x]^1].flow -= a;            x = edges[p[x]].from;        }        return a;    }    int Maxflow(int s, int t, int need)    {        this->s = s;        this->t = t;        int flow = 0;        BFS();        memset(num, 0, sizeof(num));        REP(i, n) num[d[i]]++;        int x = s;        memset(cur, 0, sizeof(cur));        while(d[s] < n)        {            if(x == t)            {                flow += Augment();                if(flow >= need) return flow;                x = s;            }            int ok = 0;            FF(i, cur[x], G[x].size())            {                Edge& e = edges[G[x][i]];                if(e.cap > e.flow && d[x] == d[e.to] + 1)   // Advance                {                    ok = 1;                    p[e.to] = G[x][i];                    cur[x] = i; // 注意                    x = e.to;                    break;                }            }            if(!ok)   // Retreat            {                int m = n-1; // 初值注意                REP(i, G[x].size())                {                    Edge& e = edges[G[x][i]];                    if(e.cap > e.flow)                        m = min(m, d[e.to]);                }                if(--num[d[x]] == 0)                    break;                num[d[x] = m + 1]++;                cur[x] = 0; // 注意                if(x != s)                    x = edges[p[x]].from;            }        }        return flow;    }    vector<int> Mincut()   // call this after maxflow    {        BFS();        vector<int> ans;        REP(i, edges.size())        {            Edge& e = edges[i];            if(!vis[e.from] && vis[e.to] && e.cap > 0)                ans.push_back(i);        }        return ans;    }    void Reduce()    {        REP(i, edges.size())            edges[i].cap -= edges[i].flow;    }    void print()    {        printf("Graph:\n");        REP(i, edges.size())            printf("%d->%d, %d, %d\n", edges[i].from, edges[i].to , edges[i].cap, edges[i].flow);    }} mf;int main(){    int n, m, x, a;    while (~RII(n, m))    {        mf.ClearAll(n + m + 2);        FE(i, 1, n)        {            mf.AddEdge(0, i, 1);            RI(x);            REP(j, x)            {                RI(a);                mf.AddEdge(i, n + a, 1);            }        }        FE(i, n + 1, m + n)            mf.AddEdge(i, n + m + 1, 1);        WI(mf.Maxflow(0, n + m + 1, INF));    }    return 0;}