传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1083

Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11353    Accepted Submission(s): 5326

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
 

题意概括:

给出 P 个课程 N 个学生 ,和选第 i 个课程的学生。询问是否所有课程都能被学生匹配,上每门课的学生都不同。

解题思路:

二分图最大匹配(匈牙利算法),判断最大匹配数是否等于 课程数 P。

AC code:

 #include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAXN = ;
const int MAXP = ; struct Edge
{
int v, nxt;
}edge[MAXP*MAXN];
int head[MAXN], cnt;
int linker[MAXN];
bool used[MAXN];
int N, P; void add(int from, int to)
{
edge[cnt].v = to;
edge[cnt].nxt = head[from];
head[from] = cnt++;
} bool Find(int x)
{
int v;
for(int i = head[x]; i != -; i = edge[i].nxt){
v = edge[i].v;
if(used[v]) continue;
used[v] = true;
if(linker[v] == - || Find(linker[v])){
linker[v] = x;
return true;
}
}
return false;
} void init()
{
memset(head, -, sizeof(head));
memset(linker, -, sizeof(linker));
memset(edge, , sizeof(edge));
cnt = ;
} int main()
{
int T_case, bnum, v;
scanf("%d", &T_case);
while(T_case--){
init();
scanf("%d%d", &P, &N);
for(int i = ; i <= P; i++){
scanf("%d", &bnum);
while(bnum--){
scanf("%d", &v);
add(i, v);
}
}
int res = ;
for(int i = ; i <= P; i++){
memset(used, , sizeof(used));
if(Find(i)) res++;
}
if(res == P) puts("YES");
else puts("NO");
}
return ;
}

HDU 1083 Courses 【二分图完备匹配】的更多相关文章

  1. 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 ...

  2. hdu 1083 Courses(二分图最大匹配)

    题意: P门课,N个学生.     (1<=P<=100    1<=N<=300) 每门课有若干个学生可以成为这门课的代表(即候选人). 又规定每个学生最多只能成为一门课的代 ...

  3. 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 ...

  4. 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 ...

  5. HDU(3605),二分图多重匹配

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3605 Escape Time Limit: 4000/2000 MS (Java/Others)    ...

  6. 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 <= ...

  7. The Accomodation of Students HDU - 2444(判断二分图 + 二分匹配)

    The Accomodation of Students Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ( ...

  8. 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 ...

  9. hdu 1083 Courses (最大匹配)

    CoursesTime Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Su ...

随机推荐

  1. Python 字典的取值

    不能用.取值 .是获取属性或方法 只能用中括号或者get方法 中括号和get中可以放字符串或者变量 get与[]的区别在于当key不存在,get不会报错,而且get可以设置取不到值时返回的默认值.

  2. BJFU 1551 ——delightful world——————【暴搜】

    delightful world 时间限制(C/C++):20000MS/30000MS          运行内存限制:65536KByte总提交:33            测试通过:10 描述 ...

  3. ASP .NET SVN && emmet 插件

    学习  ASP .NET 时间的第三周: 来讲讲如何在 visual studio 2013...上搭载 SVN吧: 废话不多说: One Step: 电脑上已安装 visual studio 201 ...

  4. js中的Function和Object

    说到构造器(condtructor).原型链(prototype),说道Function与Object,总要祭出下面这张图 1.Function是最顶层的构造器,Object是最顶层的对象 2.先有的 ...

  5. TFS2013 设置签出独占锁(转载)

    作者:晓菜鸟 出处:http://www.cnblogs.com/52XF/p/4239056.html 在使用TFS进行源代码管理的时候VS默认允许多个签出,但在团队开发中往往需要设置独占锁(排他锁 ...

  6. 2017年11月4日 vs类和结构的区别&哈希表&队列集合&栈集合&函数

    类和结构的区别 类: 类是引用类型在堆上分配,类的实例进行赋值只是复制了引用,都指向同一段实际对象分配的内存 类有构造和析构函数 类可以继承和被继承 结构: 结构是值类型在栈上分配(虽然栈的访问速度比 ...

  7. Zookeeper之集群搭建(Linux)

    Zookeeper集群搭建(Linux环境) 条件准备:准备三台Linux服务器 vt-serv1.vt-serv2.vt-serv3(虚拟机/物理机均可,服务器数量一定要是单数,不要问我为什么,据说 ...

  8. 数组和矩阵(2)——Reshape the Matrix

    In MATLAB, there is a very useful function called 'reshape', which can reshape a matrix into a new o ...

  9. mysql四-1:单表查询

    一.单表查询的语法 SELECT 字段1,字段2... FROM 表名 WHERE 条件 GROUP BY field HAVING 筛选 ORDER BY field LIMIT 限制条数 二.关键 ...

  10. 什么是SQL注入?什么是XSS攻击?什么是CSRF攻击?

    1. XSS(Cross Site Script,跨站脚本攻击) 是向网页中注入恶意脚本在用户浏览网页时在用户浏览器中执行恶意脚本的攻击方式. 1.1跨站脚本攻击分有两种形式: 反射型攻击(诱使用户点 ...