poj 1274 The Perfect Stall（匈牙利算法模板）

The Perfect Stall

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22896 Accepted: 10202

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 52 2 53 2 3 42 1 53 1 2 51 2 

Sample Output

4

题意：n头奶牛，m个地点，奶牛有自己钟爱的地点，问最多能有几头奶牛可以到自己钟爱的地点

思路：最大匹配的模板题，可以用匈牙利算法，也可以用网络流。。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;#define N 250int line[N],vis[N];///line当前与y结点相连的x的编号，vis防止成环,记录y结点访问情况int ma[N][N];int n,m;int can(int t){    for(int i=1;i<=m;i++)    {        if(!vis[i]&&ma[t][i])        {            vis[i]=1;            if(line[i]==-1||can(line[i]))            {                line[i]=t;                return 1;            }        }    }    return 0;}int main(){    int s,t;    while(~scanf("%d %d",&n,&m))    {        memset(line,-1,sizeof(line));        memset(ma,0,sizeof(ma));        for(int i=1;i<=n;i++)        {            scanf("%d",&s);            while(s--)            {                scanf("%d",&t);                ma[i][t]=1;            }        }        int ans=0;        for(int i=1;i<=n;i++)        {            memset(vis,0,sizeof(vis));            if(can(i)) ans++;        }        printf("%d\n",ans);    }    return 0;}