zoukankan      html  css  js  c++  java
  • POJ1274(二分图最大匹配)

    The Perfect Stall
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 23356   Accepted: 10405

    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

    比较裸的二分图最大匹配。
    #include <cstdio>
    #include <vector>
    #include <cstring>
    using namespace std;
    const int MAXN=405;
    vector<int> arc[MAXN];
    int n,m;
    int match[MAXN],vis[MAXN];
    bool dfs(int u)
    {
        for(int i=0;i<arc[u].size();i++)
        {
            int to=arc[u][i];
            if(!vis[to])
            {
                vis[to]=1;
                int w=match[to];
                if(w==-1||(dfs(w)))
                {
                    match[to]=u;
                    match[u]=to;
                    return true;
                }
            }    
        }
        return false;
    }
    int max_flow()
    {
        int ans=0;
        memset(match,-1,sizeof(match));
        for(int i=1;i<=n;i++)
        {
            if(match[i]==-1)
            {
                memset(vis,0,sizeof(vis));
                if(dfs(i))    ans++;
            }
        }
        return ans;
    }
    int main()
    {
        while(scanf("%d%d",&n,&m)!=EOF)
        {
            for(int i=1;i<=n;i++)    arc[i].clear();
            for(int i=1;i<=n;i++)
            {
                int s;
                scanf("%d",&s);
                for(int j=0;j<s;j++)
                {
                    int u;
                    scanf("%d",&u);    
                    u+=n;
                    arc[i].push_back(u);
                    arc[u].push_back(i);
                }
            }
            int res=max_flow();
            printf("%d
    ",res);
        }
        return 0;
    }
  • 相关阅读:
    【分布计算环境学习笔记】4 Enterprise Java Bean
    超详细的秒杀架构设计,运维,了解一下【转】
    Redis的监控指标【转】
    Windows netstat 查看端口、进程占用 查看进程路径
    wireshark抓包新手使用教程【转】
    关于设置sftp 指定端口【转】
    简单聊一聊Ansible自动化运维【转】
    tomcat启动报错SEVERE: Exception loading sessions from persistent storage【转】
    彻底搞懂 Kubernetes 的底层网络,看这几张图就够了【转】
    Java设计模式之(五)——代理模式
  • 原文地址:https://www.cnblogs.com/program-ccc/p/5837379.html
Copyright © 2011-2022 走看看