Hdu1083 Courses
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:
3 3
3 1 2 3
2 1 2
1 1
3 3
2 1 3
2 1 3
1 1
NO
——————————————————————————————
题目的意思是给出n个学生想选的课标号,每个人只能选1门课,每门课只能被1个人选,问最大有几组配对关系
思路:将学生和他想选的课建边,然后跑最大二分图匹配
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>
using namespace std; #define LL long long
const int INF = 0x3f3f3f3f;
const int MAXN=1005;
int uN,vN; //u,v数目
int g[MAXN][MAXN];
int linker[MAXN];
bool used[MAXN];
int color[MAXN]; bool dfs(int u)
{
int v;
for(v=1; v<=vN; v++)
if(g[u][v]&&!used[v])
{
used[v]=true;
if(linker[v]==-1||dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
return false;
} int hungary()
{
int res=0;
int u;
memset(linker,-1,sizeof(linker));
for(u=1; u<=uN; u++)
{
memset(used,0,sizeof(used));
if(dfs(u)) res++;
}
return res;
} int main()
{
int m,n,x,y,T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(g,0,sizeof g);
for(int i=1; i<=n; i++)
{
scanf("%d",&x);
for(int j=0;j<x;j++)
{
scanf("%d",&y);
g[i][y]=1;
}
}
uN=n,vN=m;
printf("%s\n",hungary()==n?"YES":"NO");
}
return 0;
}
Hdu1083 Courses的更多相关文章
- 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 ...
- HDU-1083 Courses 二分图 最大匹配
题目链接:https://cn.vjudge.net/problem/HDU-1083 题意 有一些学生,有一些课程 给出哪些学生可以学哪些课程,每个学生可以选多课,但只能做一个课程的代表 问所有课能 ...
- hdu2063+hdu1083(最大匹配数)
传送门:hdu2063过山车 #include <cstdio> #include <cstring> #include <string> #include < ...
- HDU1083 :Courses(二分图匹配)
Cources Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
- poj 2239 Selecting Courses (二分匹配)
Selecting Courses Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8316 Accepted: 3687 ...
- POJ 1469 COURSES
COURSES Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 20478 Accepted: 8056 Descript ...
- HDOJ 1083 Courses
Hopcroft-Karp算法模板 Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (J ...
- Courses
Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
- HDU-----(1083)Courses(最大匹配)
Courses Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total S ...
随机推荐
- 看懂 Fiddler 的瀑布图
最近准备给组内的新同学们分享下 Fiddler 这枚神器,可以讲的地方太多,我打算把一节课讲不完的内容写在博客上,大家可以随便看看.今天先介绍下 Fiddler 的瀑布图. 每个网络请求都会经历域名解 ...
- GCC编译的几个步骤
参考资料: https://blog.csdn.net/czg13548930186/article/details/78331692 一个C/C++文件要经过预处理(preprocessing).编 ...
- [RF]怎样用Robot Framework写好Test Case?
1.介绍 这是一个关于如何用Robot Framework写好Test Case的高层次的指导准则 怎样实际的与系统进行交互不在此文档范围内 最重要的准则是使测试用例尽可能的让熟悉此领域的人觉得简单易 ...
- PHP图片处理库Grafika详细教程
转载自51CTO 开发频道 1.图像基本处理:http://developer.51cto.com/art/201611/520928.htm 2.图像特效处理模块:http://developer. ...
- 摹客iDoc201901-2新功能点评
2019才刚刚开始,摹客团队就已经蓄势待发.马不停蹄地给大家带来了又一份惊喜.实话说,这次小摹都忍不住要点个赞!下面就赶紧带大家看看iDoc又更新了哪些新功能: 1.标注和评论融合.协作更高效 iDo ...
- js实现环形菜单效果
点击中间的圆点,会弹出环形菜单,效果图: 代码: <!DOCTYPE html> <html> <head> <meta charset="UTF- ...
- mysql cmd 无法登录
第一次折腾mysql诉苦记 版本注明: mysql 5.7.21 本地部署mysql,配置完成后(配置没有问题) cmd命令连接mysql: mysql -uroot -p 提示: ERROR 104 ...
- git舍弃文件更改
未进行任何提交,即文件更改在工作区 # filename 对应进行操作的文件名 git checkout -- filename 已用git add 命令提交,即文件更改在暂存区 # 舍弃暂存区的修改 ...
- springMVC学习 四 请求的中文乱码解决
在使用SpringMVC时,同样有前端向后端发送请求,请求参数中有中文,需要解决中文乱码问题,在Spring中也是向java web中一样,通过一个过滤器来解决中文乱码. 这个过滤器在spring-w ...
- Linux/Python学习路线
Linux: 初级阶段: 熟练掌握常用80个命令: 掌握Linux常用软件包的安装方法,如源码安装,rpm安装等: 学习安装设备驱动程序(如网卡,显卡驱动): 了解Grub/Lilo引导程序以及简单的 ...