hdu 3342 Legal or Not

Legal or Not

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2238    Accepted Submission(s): 1001

Problem Description

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many "holy cows" like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost "master", and Lost will have a nice "prentice". By and by, there are many pairs of "master and prentice". But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it's legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian's master and, at the same time, 3xian is HH's master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the "master and prentice" relation is transitive. It means that if A is B's master ans B is C's master, then A is C's master.

 

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y's master and y is x's prentice. The input is terminated by N = 0.
TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,..., N-1). We use their numbers instead of their names.

 

Output

For each test case, print in one line the judgement of the messy relationship.
If it is legal, output "YES", otherwise "NO".

 

Sample Input

3 20 11 22 20 11 00 0
 
import java.util.ArrayList;import java.util.PriorityQueue;import java.util.Scanner;/**   *  @功能Function Description:  拓扑排序算法  *  @开发环境Environment:       eclipse   *  @技术特点Technique:         利用拓扑排序判断有向图是否有回路   *  @算法                            *  @版本Version:               Scanner   *  @作者Author:                follow your dreams   *  @日期Date:                  20120830   *  @备注Notes:                 运行时间：27XXms  *  @链接：                     http://acm.hdu.edu.cn/showproblem.php?pid=3342 */  public class HD3342_1_20120831 {public static ArrayList<ArrayList<Integer>> g;public static int[] indegree;public static void main(String[] args) {Scanner sc = new Scanner(System.in);int m, n, a, b;ArrayList<Integer> temp;while(sc.hasNextInt()) {m = sc.nextInt();n = sc.nextInt();if(n == 0) {break;}g = new ArrayList<ArrayList<Integer>>();//初始化邻接表for(int i=0; i<m; i++) {g.add(new ArrayList<Integer>());}indegree = new int[m];//初始化入度数组for(int i=1; i<=n; i++) {a = sc.nextInt();b = sc.nextInt();temp = g.get(a);if(!temp.contains(b)) {temp.add(b);indegree[b] ++;}}topsort(m);}}public static void topsort(int m) {int count = 0;PriorityQueue<Integer> pq = new PriorityQueue<Integer>();int len = indegree.length;for(int i=0; i<len; i++) {//初始化优先级队列，将入度为0的进队if(indegree[i] == 0) {pq.add(i);}}int temp, tempLen;ArrayList<Integer> tempList;while(!pq.isEmpty()) {temp = pq.poll();count ++;tempList = g.get(temp);tempLen = tempList.size();for(int i=0; i<tempLen; i++) {indegree[tempList.get(i)]--;if(indegree[tempList.get(i)] == 0) {pq.add(tempList.get(i));}}}if(count == m) {//输出结果System.out.println("YES");} else {System.out.println("NO");}}}