POJ1274 The Perfect Stall

Time Limit: 1000MS

Memory Limit: 10000KB

64bit IO Format: %I64d & %I64u

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

USACO 40

很裸很普通的二分图匹配。

跑一遍匈牙利算法就好。

说起来今天上午模拟赛的时候，我明明已经两个月没碰过二分图匹配了，居然还是凭着本能流畅地打出了匈牙利的模板，当时超感动。

……虽然说后来知道那道题不能用匈牙利算……

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
const int mxn=1002;
int n,m;
vector<int>e[mxn];
bool vis[mxn];
int link[mxn];
int cnt=0;

int dfs(int s){
    int i,j;
    for(i=0;i<e[s].size();i++){
        int v=e[s][i];
        if(!vis[v]){
            vis[v]=1;
            if(!link[v] || dfs(link[v])){
                link[v]=s;
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        memset(link,0,sizeof link);
        for(int i=1;i<=n;i++)e[i].clear();
        cnt=0;
        int i,j;
        int num,to;
        for(i=1;i<=n;i++){
            scanf("%d",&num);
            for(j=1;j<=num;j++){
                scanf("%d",&to);
                e[i].push_back(to);
            }
        }
        for(i=1;i<=n;i++){
            memset(vis,0,sizeof vis);
            if(dfs(i))cnt++;
        }
        printf("%d
",cnt);
    }

    return 0;
}