HDU 1083 Courses( 匈牙利算法 )
来源:互联网 发布:单片机串行通信 编辑:程序博客网 时间:2024/05/20 14:16
Courses
Time Limit: 20000/10000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2915 Accepted Submission(s): 1368
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:
. 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
23 33 1 2 32 1 21 13 32 1 32 1 31 1
Sample Output
YESNO
题意:有p个课程和n个学生,每个学生可以自由选择课程(0到p个),现在要建立一个委员会,问是否能找到每个课程都有学生代表的集合,一个学生只能代表一个课程
import java.io.*;import java.util.*;public class Main {int t,p,n;int map[][]=new int[1000][3000];//记录选课情况int mark[]=new int[3000];//标记学生是否选课int link[]=new int[3000];//记录学生选的课public static void main(String[] args) {new Main().work();}void work(){Scanner sc=new Scanner(new BufferedInputStream(System.in));t=sc.nextInt();while(t--!=0){p=sc.nextInt();n=sc.nextInt();//初始化选课for(int i=1;i<=p;i++){Arrays.fill(map[i],0);int m=sc.nextInt();for(int j=1;j<=m;j++){int h=sc.nextInt();map[i][h]=1;}}int ans=0;Arrays.fill(link,0);for(int i=1;i<=p;i++){Arrays.fill(mark,0);if(DFS(i))ans++;}if(ans==p)System.out.println("YES");elseSystem.out.println("NO");}}boolean DFS(int x){for(int i=1;i<=n;i++){if(map[x][i]==1&&mark[i]==0){mark[i]=1;if((link[i]==0)||DFS(link[i])){link[i]=x;return true;}}}return false;}}
- HDU 1083 Courses( 匈牙利算法 )
- HDU-1083 Courses(匈牙利算法)
- 【匈牙利算法】hdu 1083 Courses
- hdu 1083 Courses(二分图匹配 匈牙利算法)
- HDU 1083 Courses(二分图最大匹配【匈牙利算法】)
- HDU 1083 Courses(二分图,匈牙利算法)
- hdu Courses 1083 二分匹配 ,匈牙利算法。。水题
- hdu 1083 Courses(二分匹配之匈牙利算法)
- hdu 1082 Courses 二分图最大匹配(匈牙利算法)
- HDU 1082 COURSES (二分图匹配之匈牙利算法)
- HDU 1083 Courses(二分匹配匈牙利算法模板题啊)
- HDU 1083 Courses 匈牙利算法二分匹配(邻接矩阵存关系)
- 【二分图匹配-匈牙利算法】Courses HDU
- hdu1083——Courses(匈牙利算法)
- poj_1469 COURSES匈牙利算法
- 匈牙利算法Courses
- Courses 【匈牙利算法】
- 小白算法练习 hdu courses 1083 匈牙利算法 dfs 邻接矩阵 vector模拟邻接表
- 20130830可注册域名列表(汇总)
- android自定义视图属性(atts.xml,TypedArray)学习
- Ubuntu下配置tftp服务和NFS服务
- 匈牙利算法原理讲解
- Gray code 的学习
- HDU 1083 Courses( 匈牙利算法 )
- AppDelegate
- linux守护进程编写步骤
- UVa 10881 Piotr's Ants (等价转化思想)
- 判断文件结束的三种方法(经典)
- MySQL启动方式
- C++ primer 学习记录:面对对象编程之构造函数和复制控制
- Linux下定时器的使用及实现秒以下精确定时与休眠
- POJ 2632 Crashing Robots 模拟