poj 2174 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.

二分图最大匹配模板题。

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
#define M(a) memset(a,0,sizeof(a))
vector<int> to[410];
int fa[410],v[410],m,n;
bool dfs(int x)
{
int i,j,k,u;
for (i=0;i<to[x].size();i++)
{
u=to[x][i];
if (!v[u])
{
v[u]=1;
if (!fa[u]||dfs(fa[u]))
{
fa[u]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int i,j,k,p,q,x,y,z,ans;
while (scanf("%d%d",&n,&m)==2)
{
for (i=1;i<=405;i++)
to[i].clear();
for (i=1;i<=n;i++)
{
scanf("%d",&p);
for (j=1;j<=p;j++)
{
scanf("%d",&x);
to[i].push_back(x+n);
to[x+n].push_back(i);
}
}
M(fa);
ans=0;
for (i=1;i<=n;i++)
{
M(v);
if (dfs(i)) ans++;
}
printf("%d\n",ans);
}

}