POJ1274 The Perfect Stall

                                                         The Perfect Stall
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 6632   Accepted: 3109

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

4Source

USACO 40
很明显的一个二部匹配，以前也做过这样类似的题目，是POJ1469 boys and girls。我将这个题目的代码拿来改了一下，去交，竟然超时。我觉得自己的应该是对的，才200*200怎么可能会超时，交了几次后，没办法我看了一下discuss，ym，竟然用C的输入输出会超时。我用cin和cout再交了一次，0ms。囧ing。。。。
下面是我的代码：
#include<string.h>
#include<iostream>
using namespace std;
int d[201][201],v[201],m[201],p,n;
bool dfs(int k)
{
     for(int i=1;i<=n;i++)
          if(!v[i]&&d[k][i])
          {
               v[i]=1;
               if(m[i]==0||dfs(m[i]))
               {
                    m[i]=k;
                    return true;
               }
          }
     return false;
}
int main()
{
    //int ncase;
    //scanf("%d",&ncase);
    while(cin>>p>>n)
    {
         int i,j,temp,a;
         //scanf("%d%d",&p,&n);
         memset(d,0,sizeof(d));
         memset(m,0,sizeof(m));
         for(i=1;i<=p;i++)
         {
              cin>>temp;
              for(j=1;j<=temp;j++)
              {
                    cin>>a;
                    d[i][a]=1;
              }
         }
         int sum=0;
         for(i=1;i<=p;i++)
         {
               memset(v,0,sizeof(v));          
               if(dfs(i))  sum++;
         }
         cout<<sum<<endl;
    }
    return 0;
}