click here ~~

                                   ***The Perfect Stall***

Description

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and, of course, a cow may be only assigned to one stall.

Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible. 

Input

The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will be integers in the range (1..M), and no stall will be listed twice for a given cow.

Output

For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made. 

Sample Input

5 5

2 2 5

3 2 3 4

2 1 5

3 1 2 5

1 2 

Sample Output

4

题目大意：一共同拥有n头牛，m面墙。每头牛有自己喜欢的墙，要求每堵墙仅仅能有一头牛，求最多的匹配数

解题思路：，题目要求最大的匹配数，也就是通过左边的点，从右边的点穿出。使得穿出的个数最多，每 个点仅仅能穿过一次，我们在图中加个源点和汇点即可了。

。。然后，有了源点和汇点。源点到左边每一个的流量都是 1，也就是仅仅能通过 1 次，汇点也相似，而左边的点到右 的点相应的边的边容量为 1，就这样，这道题成功转换为最大流问题。建图后最大流解决

详细看代码：

/*

Date : 2015-8-21 晚上

Author : ITAK

Motto :

今日的我要超越昨日的我。明日的我要胜过今日的我；

以创作出更好的代码为目标。不断地超越自己。

*/

#include <iostream>

#include <cstdio>

using namespace std;

///oo表示无穷大

const int oo = 1e9+5;

///mm表示边的最大数量。由于要双向建边

const int mm = 111111;

///点的最大数量

const int mn = 1000;

///node:节点数,src:源点,dest:汇点,edge:边数

int node, src, dest, edge;

///ver:边指向的结点,flow:边的流量,next:链表的下一条边

int ver[mm], flow[mm], next[mm];

///head:节点的链表头,work:用于算法中的暂时链表头,dis:距离

int head[mn], work[mn], dis[mn], q[mn];

///初始化

void Init(int _node, int _src, int _dest)

{

    node = _node, src = _src, dest = _dest;

    for(int i=0; i<node; i++)

        head[i] = -1;

    edge = 0;

}

///添加边

void addedge(int u, int v, int c)

{

    ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++;

    ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++;

}

///广搜计算出每一个点与源点的最短距离，假设不能到达汇点说明算法结束

bool Dinic_bfs()

{

    int i, u, v, l, r = 0;

    for(i=0; i<node; i++)

        dis[i] = -1;

    dis[q[r++]=src] = 0;

    for(l=0; l<r; l++)

        for(i=head[u=q[l]]; i>=0; i=next[i])

            if(flow[i] && dis[v=ver[i]]<0)

            {

                ///这条边必须有剩余流量

                dis[q[r++]=v] = dis[u] + 1;

                if(v == dest)

                    return 1;

            }

    return 0;

}

///寻找可行流的增广路算法。按节点的距离来找。加高速度

int Dinic_dfs(int u, int exp)

{

    if(u == dest)

        return exp;

    ///work 是暂时链表头，这里用 i 引用它，这样寻找过的边不再寻找*

    for(int &i=work[u],v,tmp; i>=0; i=next[i])

    {

        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)

        {

            ///正反向边容量改变

            flow[i] -= tmp;

            flow[i^1] += tmp;

            return tmp;

        }

    }

    return 0;

}

///求最大流。直到没有可行流

int Dinic_flow()

{

    int i, ret=0, data;

    while(Dinic_bfs())

    {

        for(i=0; i<node; i++)

            work[i] = head[i];

        while(data = Dinic_dfs(src, oo))

            ret += data;//cout<<666<<endl;

    }

    return ret;

}

int main()

{

    int n, m, u, v, c;

    while(cin>>n>>m)

    {

        Init(n+m+2, 0, n+m+1);

        for(u=1; u<=n; u++)

        {

            addedge(src, u, 1);

            cin>>c;

            while(c--)

            {

                cin>>v;

                addedge(u, v+n, 1);

            }

        }

        while(m)

            addedge(n+m--, dest, 1);

        cout<<Dinic_flow()<<endl;

    }

    return 0;

}