解决报告

http://blog.csdn.net/juncoder/article/details/38136193

id=1274">题目传送门

题意:

n头m个机器,求最大匹配。

ps

一分钟前刚做了POJ1469直接改了输入输出就交了,题意全然一样,,,sad ,代码传送门

The Perfect Stall
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 18108   Accepted: 8227

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

版权声明:本文博主原创文章,博客,未经同意不得转载。

POJ1274_The Perfect Stall(二部图最大匹配)的更多相关文章

  1. POJ1274 The Perfect Stall[二分图最大匹配]

    The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23911   Accepted: 106 ...

  2. POJ1274 The Perfect Stall[二分图最大匹配 Hungary]【学习笔记】

    The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23911   Accepted: 106 ...

  3. poj--1274--The Perfect Stall(最大匹配)

    The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 21665   Accepted: 973 ...

  4. poj1274 The Perfect Stall (二分最大匹配)

    Description Farmer John completed his new barn just last week, complete with all the latest milking ...

  5. [POJ] 1274 The Perfect Stall(二分图最大匹配)

    题目地址:http://poj.org/problem?id=1274 把每个奶牛ci向它喜欢的畜栏vi连边建图.那么求最大安排数就变成求二分图最大匹配数. #include<cstdio> ...

  6. Luogu 1894 [USACO4.2]完美的牛栏The Perfect Stall / POJ 1274 The Perfect Stall(二分图最大匹配)

    Luogu 1894 [USACO4.2]完美的牛栏The Perfect Stall / POJ 1274 The Perfect Stall(二分图最大匹配) Description 农夫约翰上个 ...

  7. POJ1274:The Perfect Stall(二分图最大匹配 匈牙利算法)

    The Perfect Stall Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17895   Accepted: 814 ...

  8. USACO Section 4.2 The Perfect Stall(二分图匹配)

    二分图的最大匹配.我是用最大流求解.加个源点s和汇点t:s和每只cow.每个stall和t 连一条容量为1有向边,每只cow和stall(that the cow is willing to prod ...

  9. usaco training 4.2.2 The Perfect Stall 最佳牛栏 题解

    The Perfect Stall题解 Hal Burch Farmer John completed his new barn just last week, complete with all t ...

随机推荐

  1. STL之Vector(不定长数组)

    vector是同一种对象的集合,每一个对象都有一个相应的整数索引值.和string对象一样,标准库将负责管理与存储元素相关的类存. 引入头文件 #include<vector> 1.vec ...

  2. 【POJ3159】Candies 裸的pqspfa模版题

    不多说了.就是裸的模版题. 贴代码: <span style="font-family:KaiTi_GB2312;font-size:18px;">#include & ...

  3. AppWidget应用(二)---PendingIntent 之 getActivity

    通过AppWidget应用(一)的介绍,我们已经知道如何创建一个在主界面上显示一个appWidget窗口,但这并不是我们的目的,我们需要做到程序与用户之间进行交互:下面来介绍下如何通过appWidge ...

  4. hdu1507--二分图最大匹配

    题意:你大爷.哦不! 你大叔继承了一块地什么的都是废话..,这里说说题意,和怎么建图. 题意:这里有一块N*M的地,可是有 K 个地方.是池塘,然后输入K行(x,y),OK,如今能够出售的地必须是 1 ...

  5. WPF界面设计技巧(10)-样式的继承

    原文:WPF界面设计技巧(10)-样式的继承 PS:现在我的MailMail完工了,进入内测阶段了,终于可以腾出手来写写教程了哈,关于MailMail的介绍及内测程序索取:http://www.cnb ...

  6. TCP closing a connection

    client closes socket: clientSocket.close(); step1 :client sends TCP FIN control segment to server st ...

  7. Android DES加密的CBC模式加密解密和ECB模式加密解密

    DES加密共有四种模式:电子密码本模式(ECB).加密分组链接模式(CBC).加密反馈模式(CFB)和输出反馈模式(OFB). CBC模式加密: import java.security.Key; i ...

  8. session与cookie的差别

    session     session 的工作机制是:为每一个訪客创建一个唯一的 id (UID),并基于这个 UID 来存储变量.UID 存储在 cookie 中,或者通过 URL 进行传导.   ...

  9. UVA 11427 - Expect the Expected(概率递归预期)

    UVA 11427 - Expect the Expected 题目链接 题意:玩一个游戏.赢的概率p,一个晚上能玩n盘,假设n盘都没赢到总赢的盘数比例大于等于p.以后都不再玩了,假设有到p就结束 思 ...

  10. 开源Math.NET基础数学类库使用(17)C#计算矩阵条件数

    原文:[原创]开源Math.NET基础数学类库使用(17)C#计算矩阵条件数                本博客所有文章分类的总目录:http://www.cnblogs.com/asxinyu/p ...