最大最小環算法

題目見  POJ 3895

下面是我修改過的代碼，網上原文是：https://www.cnblogs.com/gxilizq/p/3533472.html

 

package pro.yao9_22;

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Scanner;

/**
 * 
 * @author XASW
 
1 
7 8 
3 4 
1 4 
1 3 
7 1 
2 7 
7 5 
5 6 
6 2

#1 4


1 
7 9 
3 4 
1 4 
1 5 
1 3 
7 1 
2 7 
7 5 
5 6 
6 2

5？4？
 */
public class Main {
	
	static int T,N,M;
	static int answer;
	static boolean[] visited; //記錄已訪問標誌
	static LinkedList<ArrayList<Integer>> item ; //圖的 鄰接表存儲
	static int[] vNum; //保存訪問的位置，已經訪問了幾個
	public static void main(String[] args) {
		 Scanner  sc = new Scanner(System.in);
		 T = sc.nextInt();
		 for (int t = 0; t < T; t++) {
			 answer = 0;
			 N = sc.nextInt();
			 M = sc.nextInt();
			 visited = new boolean[N+1];
			 vNum = new int[N+1];
			 item = new LinkedList<ArrayList<Integer>>();
			 for (int i = 0; i <= N; i++) {
				 item.add(new ArrayList<Integer>());
			 }
			 for (int m = 0; m < M; m++) {
				 int s = sc.nextInt();
				 int e = sc.nextInt();
				 item.get(s).add(e);
				 item.get(e).add(s);
			 }
//			 printArr(item);
			 for (int i = 1; i < item.size(); i++) {
				 if(!visited[i])
					 dfs(i,1);
			 }
			 System.out.println(answer);
		 }
		 sc.close();
	}
	
	
	private static void dfs(int s, int t) {
		visited[s] = true; 
		vNum[s] = t;
		for (int i = 0; i < item.get(s).size(); i++) {
			int ns = item.get(s).get(i);
			if(!visited[ns]) {//下一個已經訪問了，就是有環了， 這裏是沒有訪問 則繼續下一個
				dfs(ns,t+1);
			}else {
				if(vNum[s]-vNum[ns]>1 //相鄰兩個不能算做環
						&&answer<vNum[s]-vNum[ns]+1
						){
					answer = vNum[s]-vNum[ns]+1;
				}
			}
		}
//		visited[s] = false; 
		
		
	}


	private static void printArr(LinkedList<ArrayList<Integer>> item2) {
		int s=0;
		for (ArrayList<Integer> arrayList : item2) {
			System.out.print(s+"===>>>");	
			for (Integer i : arrayList) {
				System.out.print(i+",");	
			}
			System.out.println("");
			s++;
		}
	}
	
}