poj 1274

       这是一道关于二分图最大匹配的入门题，刚学习相关算法的朋友们可以用此题试试手。

       这道题我用了匈牙利算法，用邻接表存了每头牛的邻集，而不是整个图，因为匈牙利算法只需要知道每个点的邻集就好了。此外，一些书上按照深搜和广搜找增广路的不同，提出了dfs版的匈牙利算法和bfs版的匈牙利算法（详参《ACM-ICPC程序设计系列  图论及应用》P199）。下面的代码用的是dfs版本的匈牙利算法。

代码（C++）：

#include <cstdlib>
#include <iostream>
#include <vector>

#define MAX 250
using namespace std;

//#define LOCAL

vector<int> G[MAX];
int cx[MAX],cy[MAX],ans; 
bool t[MAX];

void add_edge(int a,int b)
{
    G[a].push_back(b);
}

bool dfs(int u)
{
     int i,v;
     for(i=0;i<G[u].size();i++)
     {
         v=G[u][i];
         if(!t[v])
         {
            t[v]=true;      
            if(cy[v]==0||dfs(cy[v]))
            {                                              
               cx[u]=v;
               cy[v]=u;
               return true;        
            }     
         }             
     }
     return false;
}

void Hungary(int n)
{
     int i;
     memset(cx,0,sizeof(cx));
     memset(cy,0,sizeof(cy));
     
     for(i=1;i<=n;i++)
     {
         if(cx[i]==0)
         {
             memset(t,false,sizeof(t));        
             if(dfs(i)) ans++;        
         }             
     }
}

int main(int argc, char *argv[])
{
#ifdef LOCAL
   freopen("in.txt","r",stdin);
   freopen("out.txt","w",stdout);
#endif
    int n,m,t,p,i,j;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&t);
            for(j=0;j<t;j++)
            {
                scanf("%d",&p);
                add_edge(i,p);            
            }            
        }
        ans=0;
        Hungary(n);
        printf("%d\n",ans); 
        for(i=1;i<=n;i++) G[i].clear();           
    } 
    system("PAUSE");
    return EXIT_SUCCESS;
}

题目（ http://poj.org/problem?id=1274）：

The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K

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