POJ 1274
The Perfect Stall
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 17246 | Accepted: 7872 |
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.
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
求二分图的最大匹配
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 6 using namespace std; 7 8 #define maxn 505 9 10 int n,m,ans = 0; 11 int match[maxn]; 12 bool f[maxn][maxn],vis[maxn]; 13 14 bool dfs(int u) { 15 16 17 //printf("u = %d\n",u); 18 for(int i = 1; i <= n + m; ++i) { 19 if(vis[i] || !f[u][i]) continue; 20 vis[i] = 1; 21 if(match[i] == -1 || dfs(match[i])) { 22 match[i] = u; 23 return true; 24 } 25 26 } 27 28 return false; 29 30 31 } 32 void solve() { 33 for(int i = 1; i <= n + m; ++i) match[i] = -1; 34 35 for(int i = 1; i <= n + m; ++i) { 36 memset(vis,0,sizeof(vis)); 37 if(dfs(i)) ++ans; 38 } 39 } 40 41 int main() 42 { 43 44 // freopen("sw.in","r",stdin); 45 46 while(~scanf("%d%d",&n,&m) ) { 47 memset(f,0,sizeof(f)); 48 ans = 0; 49 for(int i = 1; i <= n; ++i) { 50 int num; 51 scanf("%d",&num); 52 for(int j = 1; j <= num; ++j) { 53 int ch; 54 scanf("%d",&ch); 55 f[i][ch + n] = 1; 56 57 } 58 } 59 60 solve(); 61 62 printf("%d\n",ans); 63 } 64 65 return 0; 66 67 return 0; 68 }