Courses

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5214    Accepted Submission(s): 2502

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
题解:选课,每门课程至少有一个学生;
代码:
 #include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=;
vector<int>vec[MAXN];
int usd[MAXN],vis[MAXN];
bool dfs(int x){
for(int i=;i<vec[x].size();i++){
int v=vec[x][i];
if(!vis[v]){
vis[v]=;
if(!usd[v]||dfs(usd[v])){
usd[v]=x;return true;
}
}
}
return false;
}
int main(){
int T,P,N,t,a;
scanf("%d",&T);
while(T--){
mem(usd,);
mem(vis,);
for(int i=;i<MAXN;i++)vec[i].clear();
scanf("%d%d",&P,&N);
for(int i=;i<=P;i++){
scanf("%d",&t);
while(t--){
scanf("%d",&a);
vec[i].push_back(a);
}
}
int ans=;
for(int i=;i<=P;i++){
mem(vis,);
if(dfs(i))ans++;
}
if(ans==P)puts("YES");//应该是等于P的
else puts("NO");
}
return ;
}

Courses(最大匹配)的更多相关文章

  1. hdu 1083 Courses (最大匹配)

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

  2. poj图论解题报告索引

    最短路径: poj1125 - Stockbroker Grapevine(多源最短路径,floyd) poj1502 - MPI Maelstrom(单源最短路径,dijkstra,bellman- ...

  3. HDU1083 Courses —— 二分图最大匹配

    题目链接:https://vjudge.net/problem/HDU-1083 Courses Time Limit: 20000/10000 MS (Java/Others)    Memory ...

  4. POJ1469 COURSES 【二分图最大匹配&#183;HK算法】

    COURSES Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17777   Accepted: 7007 Descript ...

  5. POJ2239 Selecting Courses(二分图最大匹配)

    题目链接 N节课,每节课在一个星期中的某一节,求最多能选几节课 好吧,想了半天没想出来,最后看了题解是二分图最大匹配,好弱 建图: 每节课 与 时间有一条边 #include <iostream ...

  6. HDU-----(1083)Courses(最大匹配)

    Courses Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total S ...

  7. [POJ] 2239 Selecting Courses(二分图最大匹配)

    题目地址:http://poj.org/problem?id=2239 Li Ming大学选课,每天12节课,每周7天,每种同样的课可能有多节分布在不同天的不同节.问Li Ming最多可以选多少节课. ...

  8. POJ2239_Selecting Courses(二分图最大匹配)

    解题报告 http://blog.csdn.net/juncoder/article/details/38154699 题目传送门 题意: 每天有12节课.一周上7天,一门课在一周有多天上课. 求一周 ...

  9. POJ2239 Selecting Courses【二部图最大匹配】

    主题链接: http://poj.org/problem?id=2239 题目大意: 学校总共同拥有N门课程,而且学校规定每天上12节可,一周上7天. 给你每门课每周上的次数,和哪一天哪一节 课上的. ...

随机推荐

  1. PowerDesigner 基础使用

    建表使用基础 1.打开软件,点击create model(如下图一)or 右上角 文件→建立新模型 or 快捷键 Ctrl+N(如下图二) 2.选择要生成脚本的数据库类型(见上图二) 3.建表(图三) ...

  2. 简单的web三层架构系统【第二版】

    昨天写了 web三层架构的第一版,准确的说是三层架构的前期,顶多算是个二层架构,要慢慢完善. 第一版里,程序虽说能运行起来,但是有一个缺陷,就是里面的SQL语句,是使用的拼接字符进行执行.这样安全系数 ...

  3. 读书笔记 SQL 事务理解

    事务的ACID属性 Atomicity 原子性 每个事务作为原子单元工作(即不可以再拆分),也就是说所有数据库变动事务,要么成功要么不成功. SQL Server把每个DML或者 DDL命令都当做一个 ...

  4. sql Servers数据库基础

    1. 数据库约束包含:     ·非空约束     ·主键约束(PK) primary key constraint 唯一且不为空     ·唯一约束(UQ) unique constraint 唯一 ...

  5. eclipse中tomcat启动项目 修改java代码不重启服务

    1.双击tomcat 2.选择modules 3.选中项目点击edit 4.去掉勾.去除auto reloading enabled 的选中 ,点击OK,

  6. objective-C学习笔记(九)ARC

    ARC叫自动引用计数Automatic Reference Counting.针对堆上的对象,管理对象的创建和释放. 哪些对象受ARC管理: OC对象指针 Block指针 使用_attribute_( ...

  7. 实时消息传输协议 RTMP(Real Time Messaging Protocol)

    实时消息传输协议(RTMP)最初是由 Macromedia 为互联网上 Flash player 和服务器之间传输音频.视频以及数据流而开发的一个私有协议.Adobe 收购 Macromedia 购以 ...

  8. [LeetCode]题解(python):084-Largest Rectangle in Histogram

    题目来源: https://leetcode.com/problems/largest-rectangle-in-histogram/ 题意分析: 给定一个数组,数组的数字代表这个位置上的bar的高度 ...

  9. C++ 面向对象学习2 构造方法

    Date.h #ifndef DATE_H #define DATE_H class Date{ public: Date(,,);//自定义了构造方法 会覆盖掉默认的无参构造方法 void setD ...

  10. DC游戏《斑鸠》原创赏析[转载]

    游戏背景:      凤来之国本来只是边远地区的一个小国.但现在他们却自称为得到“神之力”的“神通者”,在“选民思想”“和平统一”之类的名义下开始了对各地的武力侵略.      事情的起因是因为凤来之 ...