题目链接:http://poj.org/problem?id=1469

COURSES
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 21229   Accepted: 8355

Description

Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is possible to form a committee of exactly P students that satisfies simultaneously the conditions:

  • every student in the committee represents a different course (a student can represent a course if he/she visits that course)
  • each course has a representative in the committee

Input

Your
program should read sets of data from the std input. The first line of
the input contains the number of the data sets. Each data set is
presented in the following format:

P N

Count1 Student1 1 Student1 2 ... Student1 Count1

Count2 Student2 1 Student2 2 ... Student2 Count2

...

CountP StudentP 1 StudentP 2 ... StudentP CountP

The first line in each data set contains two positive integers
separated by one blank: P (1 <= P <= 100) - the number of courses
and N (1 <= N <= 300) - the number of students. The next P lines
describe in sequence of the courses �from course 1 to course P, each
line describing a course. The description of course i is a line that
starts with an integer Count i (0 <= Count i <= N) representing
the number of students visiting course i. Next, after a blank, you抣l
find the Count i students, visiting the course, each two consecutive
separated by one blank. Students are numbered with the positive integers
from 1 to N.

There are no blank lines between consecutive sets of data. Input data are correct.

Output

The
result of the program is on the standard output. For each input data set
the program prints on a single line "YES" if it is possible to form a
committee and "NO" otherwise. There should not be any leading blanks at
the start of the line.

Sample Input

2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1

Sample Output

YES
NO

Source

 
题意:

给你p门课程和n个学生,一个学生可以选0门,1门,或者多门课程,现在要求一个由p个学生组成的集合,满足下列2个条件:

1.每个学生选择一个不同的课程

2.每个课程都有不同的代表

如果满足,就输出YES

分析:

做最大匹配,要是匹配数是P就是YES.

#include <stdio.h>
#include <string.h> int p,n;
bool maps[][];
bool use[];
int match[]; bool DFS(int u)
{
for(int i=; i<=n; i++)
{
if(!use[i]&&maps[u][i])
{
use[i] = true;
if(match[i]==-||DFS(match[i]))
{
match[i] = u;
return true;
}
}
}
return false;
} int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(match,-,sizeof(match));
memset(maps,false,sizeof(maps));
scanf("%d%d",&p,&n);
int num = ;
for(int i=; i<=p; i++)
{
int stu;
scanf("%d",&stu);
for(int j=; j<=stu; j++)
{
int v;
scanf("%d",&v);
maps[i][v] = true;
}
} for(int i=; i<=p; i++)
{ memset(use,false,sizeof(use));
if(DFS(i))
num++;
}
if(num==p)
printf("YES\n");
else printf("NO\n"); } return ;
}

Poj(1469),二分图最大匹配的更多相关文章

  1. poj 1469 二分图最大匹配

    就是最简单的最大匹配,没的说 #include<iostream> #include<cstdio> #include<cstring> #include<a ...

  2. poj 1469(二分图 最大匹配)

    这道题让我认识到了c++cin,cout确实会使其超时,还是我用的c printf吧 #include<cstdio> #include<iostream> #include& ...

  3. poj 2239 二分图最大匹配,基础题

    1.poj 2239   Selecting Courses   二分图最大匹配问题 2.总结:看到一个题解,直接用三维数组做的,很巧妙,很暴力.. 题意:N种课,给出时间,每种课在星期几的第几节课上 ...

  4. POJ 2226二分图最大匹配

    匈牙利算法是由匈牙利数学家Edmonds于1965年提出,因而得名.匈牙利算法是基于Hall定理中充分性证明的思想,它是二部图匹配最常见的算法,该算法的核心就是寻找增广路径,它是一种用增广路径求二分图 ...

  5. POJ Evacuation /// 二分图最大匹配

    题目大意: 在一个n*m的房间中 ‘X’为墙 ‘D’为门 ‘.’为人 门只存在与外围 人每秒钟只能向四连通区域走一步 门比较狭窄 每秒钟只能通过一个人 求所有人逃脱的最短时间 如果不可能则输出impo ...

  6. poj 2724 二分图最大匹配

    题意: 会给出M个串,我们要做的就是将这M个串给清除了.对于任意两个串,若二进制形式只有一位不一样,那么这两个串可以在一次操作消除,否则每个操作只能消除一个串. 3 3 *01 100 011 可以代 ...

  7. Asteroids - poj 3041(二分图最大匹配问题)

      Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 17258   Accepted: 9386 Description Be ...

  8. poj 2446 二分图最大匹配

    思路:由(i+j)为偶数的点向(i+j)为奇数的点建边.求一次最大匹配,若正好为空格数(不包含洞)的一半,即输出YES. #include<iostream> #include<cs ...

  9. POJ 1719 二分图最大匹配(记录路径)

    Shooting Contest Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 4097   Accepted: 1499 ...

  10. poj 3692 二分图最大匹配

    思路: 如果我们将认识的建边,求最大独立集就是互相不认识的人数.那么我们反过来,将不认识的建图,求最大独立集就是互相认识的人数. #include<cstdio> #include< ...

随机推荐

  1. 使用Mysql修改密码命令更改root的密码

    使用Mysql修改密码命令更改root的密码. 进入Mysql数据库命令行方式有两种方式: 方式一:在Mysql开始菜单里包含Mysql命令行客户端,只要点击输入root的密码即可进入. 方式二:在D ...

  2. Andriod环境搭建

    安卓是一款现在在移动端十分流行的系统,本人出于好奇心,希望彻底了解安卓的开发技. 首先了解一下安卓的系统构架,安卓大致分为四层架构,五块区域: 1.Linux内核层 Andriod是基于Linux2. ...

  3. 夺命雷公狗---Thinkphp----9之中间层的创建,防止跨目录访问

    我们创建一个CommonController.class.php的中间层,让后让别的控制器都直接继承CommonController这个控制器即可决解跨目录访问的问题, <?php namesp ...

  4. Android真机测试 INSTALL_FAILED_INSUFFICIENT_STORAGE 解决方法[转]

    方法一: 试试修改一下manifest文件 :添加 一句:   android:installLocation="preferExternal" [html]view plainc ...

  5. zw版【转发·台湾nvp系列Delphi例程】HALCON CropPart

    zw版[转发·台湾nvp系列Delphi例程]HALCON CropPart procedure TForm1.Button1Click(Sender: TObject);var ho_Egypt1 ...

  6. 【Install】我是如何安装Linux类系统的

    安装系统:ubuntu12.04 i386 DVD U盘启动12.04live系统   连线,设置连接 安装系统到硬盘   “语言支持”,更新   安装gnome经典界面 sudo apt-get i ...

  7. 「ruby」使用rmagick处理图像

    安装rmagick gem A new release 2.13.2 of RMagick is now available on github as well as rubygems. This r ...

  8. C++之: CDib类

    头文件Cdib.h 源文件Cdib.cpp

  9. RMQ(非log2储存方法)

    2016-03-31 RMQ 难度级别:B: 运行时间限制:1000ms: 运行空间限制:256000KB: 代码长度限制:2000000B 试题描述 长度为n的数列A,以及q个询问,每次询问一段区间 ...

  10. python paramiko ssh.exec_command()启动tomcat服务器应用进程失败问题解决方法- Neither the JAVA_HOME nor the JRE_HOME environment variable is defined At least one of these environment variable is needed to run this progr

    问题说明: