HUD——1083 Courses
HUD——1083 Courses
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7699 Accepted Submission(s):
3781
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
Your program should read sets of data from a text file. The
first line of the input file 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'll 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.
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.
An example of program
input and output:
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define N 501
using namespace std;
bool vis[N];
int n,m,x,y,T,t,ans,pre[N],map[N][N];
int read()
{
,f=;char ch=getchar();
; ch=getchar();}
+ch-'; ch=getchar();}
return x*f;
}
int find(int x)
{
;i<=n;i++)
{
if(!vis[i]&&map[x][i])
{
vis[i]=true;
||find(pre[i]))
{
pre[i]=x;
;
}
}
}
;
}
int main()
{
T=read();
while(T--)
{
ans=;
memset(map,,sizeof(map));
m=read(),n=read();
;i<=m;i++)
{
t=read();
while(t--)
{x=read();map[i][x]=;}
}
if(n<m) printf("NO\n");
else
{
memset(pre,-,sizeof(pre));
;i<=m;i++)
{
memset(vis,,sizeof(vis));
if(find(i)) ans++;
}
// printf("%d\n",ans);
if(ans==m) printf("YES\n");
else printf("NO\n");
}
}
;
}
HUD——1083 Courses的更多相关文章
- HDU 1083 Courses 【二分图完备匹配】
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1083 Courses Time Limit: 20000/10000 MS (Java/Others) ...
- HDU 1083 - Courses - [匈牙利算法模板题]
题目链接:http://acm.split.hdu.edu.cn/showproblem.php?pid=1083 Time Limit: 20000/10000 MS (Java/Others) M ...
- HDOJ 1083 Courses
Hopcroft-Karp算法模板 Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...
- hdoj 1083 Courses【匈牙利算法】
Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
- hdu - 1083 - Courses
题意:有P门课程,N个学生,每门课程有一些学生选读,每个学生选读一些课程,问能否选出P个学生组成一个委员会,使得每个学生代言一门课程(他必需选读其代言的课程),每门课程都被一个学生代言(1 <= ...
- HDU - 1083 Courses /POJ - 1469
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1083 http://poj.org/problem?id=1469 题意:给你P个课程,并且给出每个课 ...
- HDU 1083 Courses(二分图匹配模板)
http://acm.hdu.edu.cn/showproblem.php?pid=1083 题意:有p门课和n个学生,每个学生都选了若干门课,每门课都要找一个同学来表演,且一个同学只能表演一门课,判 ...
- hdu 1083 Courses (最大匹配)
CoursesTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- HDU 1083 Courses(最大匹配模版题)
题目大意: 一共有N个学生跟P门课程,一个学生可以任意选一 门或多门课,问是否达成: 1.每个学生选的都是不同的课(即不能有两个学生选同一门课) 2.每门课都有一个代表(即P门课都被成功选过 ...
随机推荐
- IOS应用
下面是这个类的一些功能: 1.设置icon上的数字图标 //设置主界面icon上的数字图标,在2.0中引进, 缺省为0 [UIApplicationsharedApplication].applica ...
- 使用windows的fsutil命令创建指定大小及类型的测试文件
在软件测试中,对于上传.下载一类功能常常需要用不同大小的文件进行测试. 使用Windows命令fsutil可以生成任意大小.任意类型文件. C:\Users\axia\fsutil file crea ...
- 项目中常用git命令操作指令(一般正常的话够用不够再看相关git命令)
配置git1.首先在本地创建ssh key:ssh-keygen -t rsa -C "github上注册的邮箱" //(一路回车)2.进入c:/Users/xxxx_000/.s ...
- Hibernate Lazy属性与懒加载 整理
lazy概念:要用到的时候,再去加载,对于关联的集合来说,只有当访问到的时候,才去加载它所关联的集合,比如一个user对应很多权限,只有当user.getRights()的时候,才发出select r ...
- EOS Dawn 3.0 智能合约 -- 新格式
1.简介 随着EOS Dawn 3.0发布,智能合约的坑又要重新踩了o(╥﹏╥)o:3.0不仅将原来本身就在链里的基础合约独立出来,简单的介绍见3.0合约改变,合约的书写方式也有巨大变化,相比之前更加 ...
- RGB颜色空间、色调、饱和度、亮度,HSV颜色空间详解
本文章会详细的介绍RGB颜色空间与RGB三色中色调.饱和度.亮度之间的关系,最后会介绍HSV颜色空间! RGB颜色空间 概述 RGB颜色空间以R(Red:红).G(Green:绿).B(Blue:蓝) ...
- Zend Studio 修改“代码字体和大小”
- 数据库sql语句limit区别
注意:并非所有的数据库系统都支持 SELECT TOP 语句. MySQL 支持 LIMIT 语句来选取指定的条数数据, Oracle 可以使用 ROWNUM 来选取. SQL Server / MS ...
- java程序在一个电脑上只启动一次,只开一个进程
方案1: 单进程程序可以用端口绑定.程序启动的时候可以尝试看该端口是否已经被占用,如果占用则程序已经启动. 方案2:你可以在java程序中创建一个隐藏文件,程序退出的时候删除这个文件.这样在程序启动的 ...
- net core 使用ef生成实体类(SqlServer)
1)打开程序包管理器控制台 2)输入命令 Install-Package Microsoft.EntityFrameworkCore.SqlServer 3)输入命令 Install-Packag ...