poj The Perfect Stall 1274 （二分图最大匹配）

The Perfect Stall

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 22060 Accepted: 9872

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：

有n头牛，m种食物，每头牛都有自己喜爱吃的食物，并且每种食物只够一头牛吃，问最多有几头牛可以吃到自己喜爱的食物？

下面输入有n行，每行第一个数为第i头牛喜爱的食物数k，然后再输入k个数，表示食物的编号。

//思路：http://blog.csdn.net/yanghui07216/article/details/50976738

最大匹配问题，直接套模板。

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>#include<iostream>#define INF 0x3f3f3f3f#define IN __int64#define ull unsigned long long#define ll long long#define N 210#define M 1000000007using namespace std;int map[N][N];int x[N],y[N];int vis[N];int n,m;int dfs(int xx){for(int i=1;i<=m;i++){if(!vis[i]&&map[xx][i]){vis[i]=1;if(!y[i]||dfs(y[i])){y[i]=xx;x[xx]=i;return 1;}}}return 0;}int main(){int i,j;while(scanf("%d%d",&n,&m)!=EOF){memset(map,0,sizeof(map));memset(x,0,sizeof(x));memset(y,0,sizeof(y));for(i=1;i<=n;i++){int k;scanf("%d",&k);while(k--){scanf("%d",&j);map[i][j]=1;}}int sum=0;for(i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(dfs(i))sum++;}printf("%d\n",sum);}return 0;}