The Perfect Stall
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 17768 | Accepted: 8104 |
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
Source
模板题：
1 //168K 16MS C++ 960B 2014-06-03 12:15:26
2 #include<iostream>
3 #include<vector>
4 #define N 205
5 using namespace std;
6 vector<int>V[N];
7 int vis[N];
8 int match[N];
9 int n,m;
10 int dfs(int u)
11 {
12 for(int i=0;i<V[u].size();i++){
13 int v=V[u][i];
14 if(!vis[v]){
15 vis[v]=1;
16 if(match[v]==-1 || dfs(match[v])){
17 match[v]=u;
18 return 1;
19 }
20 }
21 }
22 return 0;
23 }
24 int hungary()
25 {
26 memset(match,-1,sizeof(match));
27 int ret=0;
28 for(int i=1;i<=n;i++){
29 memset(vis,0,sizeof(vis));
30 ret+=dfs(i);
31 }
32 return ret;
33 }
34 int main(void)
35 {
36 int k,a;
37 while(scanf("%d%d",&n,&m)!=EOF)
38 {
39 for(int i=0;i<=n;i++) V[i].clear();
40 for(int i=1;i<=n;i++){
41 scanf("%d",&k);
42 while(k--){
43 scanf("%d",&a);
44 V[i].push_back(a);
45 }
46 }
47 printf("%d\n",hungary());
48 }
49 return 0;
50 }
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