POJ 1274-The Perfect Stall（二分图匹配/最大流问题）

r囧r小猫 2022-09-25 04:22 123阅读 0赞

The Perfect Stall

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;23108</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;10297</td> 
  </tr> 
 </tbody> 
</table>

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

Source

[USACO 40][]

**题目意思：**

农民有N个奶牛，M个谷仓，每头牛只愿意在指定的谷仓内产奶。

下面N行，每行的第一个数S表示对应这头牛愿意去的谷仓总数，之后的S个数是各个谷仓的编号。

**解题思路：**

看成有向图，二分图匹配问题，奶牛和谷仓总数就是图中的顶点总数。

转换成最大流问题时，增加源点和汇点。

#include <iostream>
    #include <cstdlib>
    #include <cstring>
    #include <vector>
    #define MAXN 5000
    using namespace std;
    
    int V;//顶点数
    int R[MAXN],C[MAXN];
    vector<int>G[MAXN];//图的邻接表表示
    int match[MAXN];//所匹配的顶点
    bool used[MAXN];//DFS访问标记
    
    void add_edge(int u,int v)//向图中增加一条连接u和v的边
    {
        G[u].push_back(v);
        G[v].push_back(u);
    }
    
    bool dfs(int v)//dfs寻找增广路
    {
        used[v]=true;
        for(int i=0; i<G[v].size(); ++i)
        {
            int u=G[v][i],w=match[u];
            if(w<0||!used[w]&&dfs(w))
            {
                match[v]=u;
                match[u]=v;
                return true;
            }
        }
        return false;
    }
    
    int bipartite_matching()//二分图最大匹配
    {
        int res=0;
        memset(match,-1,sizeof(match));
        for(int v=0; v<V; ++v)
            if(match[v]<0)
            {
                memset(used,0,sizeof(used));
                if(dfs(v)) ++res;
            }
        return res;
    }
    
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int N,M;
        while(cin>>N>>M)
        {
            V=N+M;
            for(int i=0; i<V; ++i)
                G[i].clear();
            for(int i=0; i<N; ++i)
            {
                int S;
                cin>>S;
                for (int j=0; j<S; ++j)
                {
                    int u;
                    cin>>u;
                    u=N+u-1;
                    add_edge(i,u);
                }
            }
            cout<<bipartite_matching()<<endl;
        }
        return 0;
    }
    /**
    5 5
    2 2 5
    3 2 3 4
    2 1 5
    3 1 2 5
    1 2
    **/

[USACO 40]: http://poj.org/searchproblem?field=source&key=USACO+40