hdoj--1083--Courses(最大匹配)
Courses
. 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:
2
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
YES
NO
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
int pipei[500],used[500];
vector<int>map[500];
int find(int x)
{
for(int i=0;i<map[x].size();i++)
{
int y=map[x][i];
if(!used[y])
{
used[y]=1;
if(!pipei[y]||find(pipei[y]))
{
pipei[y]=x;
return 1;
}
}
}
return 0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,m;
scanf("%d%d",&n,&m);
memset(pipei,0,sizeof(pipei));
for(int i=1;i<=n;i++)
map[i].clear();
for(int i=1;i<=n;i++)
{
int x,y;
int k;
scanf("%d",&k);
while(k--)
{
scanf("%d",&y);
map[i].push_back(y);
}
}
int sum=0;
for(int i=1;i<=n;i++)
{
memset(used,0,sizeof(used));
sum+=find(i);
}
if(sum==n)
printf("YES\n");
else
printf("NO\n");
}
return 0;
}
hdoj--1083--Courses(最大匹配)的更多相关文章
- 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 (最大匹配)
CoursesTime Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Su ...
- 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) ...
- HUD——1083 Courses
HUD——1083 Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Ot ...
- 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 ...
- 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 ...
- 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 ...
- 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
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 <= ...
随机推荐
- 6月7号shiro
Retains all Cache objects maintained by this cache manager :保留此缓存管理器维护的所有缓存对象 Destroyable可毁灭的 retain ...
- C# 怎么把类文件如(XXX.cs)转为dll文件
打开VS2012或2017 ,新建项目,选择 类库(.NET Framework),创建好一个项目 在建好的项目中添加需要转的类文件 然后将项目重新生成后,在项目的Debug下就可以找到对应的dll ...
- 使用T-sql建库建表建约束
为什么要使用sql语句建库建表? 现在假设这样一个场景,公司的项目经过测试没问题后需要在客户的实际环境中进行演示,那就需要对数据进行移植,现在问题来了:客户的数据库版本和公司开发阶段使用的数据库不兼容 ...
- ANN:ML方法与概率图模型
一.ML方法分类: 产生式模型和判别式模型 假定输入x,类别标签y - 产生式模型(生成模型)估计联合概率P(x,y),因可以根据联合概率来生成样本:HMMs ...
- Linux命令小记
以下说法都是基于普通用户的角度,如果是root,可能会有不同. (1)rm -r或-R选项:递归删除目录及其内容(子目录.文件) rm默认无法删除目录,如果删除空目录,可以使用-d选项.如果目录非空, ...
- php知识点(基本上文档都有,只为方便记忆)
类型转换 (unset)转换为NULL (binary) 转换和 b 前缀转换支持为 PHP 5.2.1 新增 转换二进制 隐藏php后缀名 AddType application/x-httpd ...
- [Advanced Algorithm] - Exact Change
题目 设计一个收银程序 checkCashRegister(),其把购买价格(price)作为第一个参数 , 付款金额 (cash)作为第二个参数, 和收银机中零钱 (cid) 作为第三个参数. ci ...
- Step by Step 开发dynamics CRM
这里是作为开发贴的总结. 现在plugin和workflow系列已经终结. 希望这些教程能给想入坑的小伙伴一些帮忙. CRM中文教材不多, 我会不断努力为大家提供更优质的教程. Plugin 开发系列 ...
- BZOJ 2959: 长跑 LCT_并查集_点双
真tm恶心...... Code: #include<bits/stdc++.h> #define maxn 1000000 using namespace std; void setIO ...
- [NOI2015]软件包管理器 树链剖分_线段树
没有太大难度,刷水有益健康 Code: // luogu-judger-enable-o2 #include <bits/stdc++.h> #define setIO(s) freope ...