Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4175    Accepted Submission(s): 1990

Problem 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

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:

 
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
 
Recommend
/**
题意:n个student,m个course,然后每个course选一个课代表,然后看能否为每门course
选到一名课代表
做法:匈牙利算法
**/
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<string.h>
using namespace std;
#define maxn 310
int linker[maxn];
int un,vn;
int g[maxn][maxn];
bool used[maxn];
int n,m;
int dfs(int u)
{
for(int i=;i<=m;i++)
{
if(g[u][i] && used[i] == false)
{
used[i] = true;
if(linker[i] == - || dfs(linker[i]))
{
linker[i] = u;
return ;
}
}
}
return ;
}
int hungary()
{
int res = ;
memset(linker,-,sizeof(linker));
for(int i=;i<n;i++)
{
memset(used,false,sizeof(used));
res += dfs(i);
}
return res;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif // ONLINE_JUDGE
int T;
scanf("%d",&T);
while(T--)
{ scanf("%d %d",&n,&m);
int Q;
memset(g,,sizeof(g));
int v;
for(int i=;i<n;i++)
{
scanf("%d",&Q);
while(Q--)
{
scanf("%d",&v);
g[i][v] = ;
}
}
int res = ;
res = hungary();
if(res == n) printf("YES\n");
else printf("NO\n");
}
return ;
}

HDU-1083的更多相关文章

  1. HDU 1083 网络流之二分图匹配

    http://acm.hdu.edu.cn/showproblem.php?pid=1083 二分图匹配用得很多 这道题只需要简化的二分匹配 #include<iostream> #inc ...

  2. 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 <= ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. (匹配)Courses -- hdu --1083

    链接: http://acm.hdu.edu.cn/showproblem.php?pid=1083 http://acm.hust.edu.cn/vjudge/contest/view.action ...

  7. 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)  ...

  8. 【hdu 1083】Courses

    [Link]:http://acm.hdu.edu.cn/showproblem.php?pid=1083 [Description] 有p门的课,每门课都有若干学生,现在要为每个课程分配一名课代表, ...

  9. 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 ...

  10. C - Courses - hdu 1083(模板)

    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) 输入为: P N( ...

随机推荐

  1. BZOJ1492: [NOI2007]货币兑换Cash 【dp + CDQ分治】

    1492: [NOI2007]货币兑换Cash Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 5391  Solved: 2181 [Submit][S ...

  2. 【DP】【P2224】】【HNOI2001】产品加工

    传送门 Description 某加工厂有\(A\).\(B\)两台机器,来加工的产品可以由其中任何一台机器完成,或者两台机器共同完成.由于受到机器性能和产品特性的限制,不同的机器加工同一产品所需的时 ...

  3. 背景建模技术(四):视频分析(VideoAnalysis)模块

    视频分析模块主要包含两个函数,一个是VideoAnalysis::setup(....),其主要功能就是确定测试的视频是视频文件或摄像头输入亦或是采用命令行参数:第二个函数是VideoAnalysis ...

  4. HDU1711 KMP(模板题)

    Number Sequence Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) ...

  5. ES6中字符串的扩展

    一.查找字符串 在ES5中,可以使用 indexOf 方法和 lastIndexOf 方法查找字符串: let str = 'hello world'; alert(str.indexOf('o')) ...

  6. linux下输出查看进程及杀进程

    1.查找有关tomcat的进程 ps -ef | grep tomcat 2.查看某端口占用情况 netstat -tulpn | grep 9009 3.杀进程 普通:kill 进程id 强制:ki ...

  7. 元类编程--property动态属性

    from datetime import date, datetime class User: def __init__(self, name, birthday): self.name = name ...

  8. Spring 框架的设计理念与设计模式分析(山东数漫江湖)

    Spring 的骨骼架构 Spring 总共有十几个组件,但是真正核心的组件只有几个,下面是 Spring 框架的总体架构图: 图 1 .Spring 框架的总体架构图 从上图中可以看出 Spring ...

  9. bzoj 2730 割点

    首先我们知道,对于这张图,我们可以枚举坍塌的是哪个点,对于每个坍塌的点,最多可以将图分成若干个不连通的块,这样每个块我们可能需要一个出口才能满足题目的要求,枚举每个坍塌的点显然是没有意义的,我们只需要 ...

  10. 深入理解 JavaScript(五)

    根本没有“JSON 对象”这回事! 前言 写这篇文章的目的是经常看到开发人员说:把字符串转化为 JSON 对象,把 JSON 对象转化成字符串等类似的话题,所以把之前收藏的一篇老外的文章整理翻译了一下 ...