HDU1083 最大二分匹配
来源:互联网 发布:足球怎么过人知乎 编辑:程序博客网 时间:2024/05/17 04:27
Courses
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
import java.io.BufferedReader;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import java.util.ArrayList;import java.util.Arrays;import java.util.List;import java.util.StringTokenizer;public class Main {public static void main(String[] args) {new Task().solve();}}class Task {InputReader in = new InputReader(System.in);PrintWriter out = new PrintWriter(System.out);int[] match;boolean[] used;List<Integer> adj[];boolean dfs(int u) {for (int v : adj[u]) {if(used[v]){continue ;}used[v] = true;if (match[v] == -1 || dfs(match[v])) {match[v] = u;return true;}}return false;}int maxMatch(int n) {int sum = 0;Arrays.fill(match, -1);for (int u = 1; u <= n; u++) {Arrays.fill(used, false);if (dfs(u)) {sum++;}}return sum;}void solve() {int t = in.nextInt();while (t-- > 0) {int P = in.nextInt();int N = in.nextInt();adj = new List[P + 1];used = new boolean[N + 1];match = new int[N+1] ;for (int i = 1; i <= P; i++) {adj[i] = new ArrayList<Integer>();}for (int i = 1; i <= P; i++) {int m = in.nextInt();while (m-- > 0) {adj[i].add(in.nextInt());}}out.println(maxMatch(P) == P ? "YES" : "NO");}out.flush();}}class InputReader {public BufferedReader reader;public StringTokenizer tokenizer;public InputReader(InputStream stream) {reader = new BufferedReader(new InputStreamReader(stream), 32768);tokenizer = new StringTokenizer("");}private void eat(String s) {tokenizer = new StringTokenizer(s);}public String nextLine() {try {return reader.readLine();} catch (Exception e) {return null;}}public boolean hasNext() {while (!tokenizer.hasMoreTokens()) {String s = nextLine();if (s == null)return false;eat(s);}return true;}public String next() {hasNext();return tokenizer.nextToken();}public int nextInt() {return Integer.parseInt(next());}public long nextLong() {return Long.parseLong(next());}public double nextDouble() {return Double.parseDouble(next());}public BigInteger nextBigInteger() {return new BigInteger(next());}}
阅读全文
0 0
- HDU1083 最大二分匹配
- 【HDU1083】【最大二分匹配】
- HDU1083 最大二分匹配
- hdu1083【二分图】【最大匹配】
- HDU1083(二分图+最大匹配+匈牙利算法)
- hdu1083 Courses ( 二分图最大匹配)
- HDU1083--Courses(二分图最大匹配)
- hdu1083培训班 ~~二分图最大匹配
- 【二分匹配】 hdu1083 Courses
- hdu1083二分匹配基础
- 【二分匹配】HDU1083-Courses
- HDU1083 Courses 二分匹配
- hdu1083二分图匹配
- HDU1083+POJ1469 (匈牙利算法+最大二分匹配)
- hdu1083——二分匹配
- hdu1083 Courses(二分匹配)
- hdu1083(二分图匹配)
- Courses HDU1083 -二分图匹配
- vue--computed
- 选择排序
- MFC:工具栏不显示图标
- B 同花顺
- SQL行转列汇总
- HDU1083 最大二分匹配
- [YTU]_2535( C++复数运算符重载(+与<<))
- 在C#中单击右键添加引用时弹出错误提示对话框
- 2017年的迷茫与悲痛
- 如何用pip安装xgboost
- POJ 1927 Area in Triangle 笔记
- udelay();mdelay();ndelay();msleep();
- 第七章——图
- 为初学者介绍 JavaScript 中的 this 关键字