POJ-1274The Perfect Stall,二分匹配裸模板题

The Perfect Stall

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 23313		Accepted: 10383

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

说实话做此题并没有很认真的看题（手动反省）；

大概的意思就是有N只奶牛，M个牛圈，每个牛圈只能容下一只奶牛，而每只奶牛都有自己喜欢的牛圈，不然就不产奶。

求怎么分配使得最终牛奶产量最高。

典型的二分最大匹配，用来熟悉匈牙利算法邻接矩阵模板实现，加深对模板的熟练程度。

const int N=200+10;
int g[N][N],v[N],linked[N],n,m;
int dfs(int u)
{
    for(int i=1;i<=m;i++)
        if(!v[i]&&g[u][i])
    {
        v[i]=1;
        if(linked[i]==-1||dfs(linked[i]))
        {
            linked[i]=u;
            return 1;
        }
    }
    return 0;
}
void hungary()
{
    int ans=0;
    memset(linked,-1,sizeof(linked));
    for(int i=1;i<=n;i++)
    {
        memset(v,0,sizeof(v));
        if(dfs(i)) ans++;
    }
    printf("%d\n",ans);
}
int main()
{
    int i,x;
    while(~scanf("%d%d",&n,&m))
    {
        memset(g,0,sizeof(g));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&x);
            int p;
            while(x--)
            {
                scanf("%d",&p);
                g[i][p]=1;
            }
        }
        hungary();
    }
    return 0;
}

还是在于模型的建立上，只要确立好了模型方法自然就有了。

做二分匹配的经典步骤：理清题意—>确立模型——>选取模板