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2013
11-10

POJ 2230 Watchcow [解题报告] Java

Watchcow

问题描述 :

Bessie’s been appointed the new watch-cow for the farm. Every night, it’s her job to walk across the farm and make sure that no evildoers are doing any evil. She begins at the barn, makes her patrol, and then returns to the barn when she’s done.

If she were a more observant cow, she might be able to just walk each of M (1 <= M <= 50,000) bidirectional trails numbered 1..M between N (2 <= N <= 10,000) fields numbered 1..N on the farm once and be confident that she's seen everything she needs to see. But since she isn't, she wants to make sure she walks down each trail exactly twice. It's also important that her two trips along each trail be in opposite directions, so that she doesn't miss the same thing twice.

A pair of fields might be connected by more than one trail. Find a path that Bessie can follow which will meet her requirements. Such a path is guaranteed to exist.

输入:

* Line 1: Two integers, N and M.

* Lines 2..M+1: Two integers denoting a pair of fields connected by a path.

输出:

* Lines 1..2M+1: A list of fields she passes through, one per line, beginning and ending with the barn at field 1. If more than one solution is possible, output any solution.

样例输入:

4 5
1 2
1 4
2 3
2 4
3 4

样例输出:

1
2
3
4
2
1
4
3
2
4
1

温馨提示:

OUTPUT DETAILS:

Bessie starts at 1 (barn), goes to 2, then 3, etc…

解题代码:

//* @author: 
import java.util.*;

class Edge{
	int v, next;
	boolean vis;
	Edge(int v,int next){
		this.v = v;
		this.next = next;
		this.vis = false;
	}
}
class Graph{
	static int MAXN = 10010;
	static int n, m, size;
	static int[] ans = new int[MAXN*3];
	static int[] head = new int[MAXN];
	static Edge[] edge = new Edge[MAXN*10];
	Graph(){
		Arrays.fill(head,-1);
		size = 0;
	}
	void add_edge(int u,int v){
		edge[size] = new Edge(v,head[u]);
		head[u] = size ++;
	}

	void Find_Euler(int u){    	
       	for(int i = head[u];i != -1;i = edge[i].next){
			if(!edge[i].vis){   
				edge[i].vis = true;
				Find_Euler(edge[i].v);
			}
		}
		System.out.println(u);
	}

}
public class Main{
	public static void main(String []args){
		Scanner in = new Scanner(System.in);
		while(in.hasNext()){
			Graph a = new Graph();
			a.n = in.nextInt();
			a.m = in.nextInt();
			for(int i = 1;i <= a.m; ++i){
				int u = in.nextInt();
				int v = in.nextInt();
				a.add_edge(u,v);
				a.add_edge(v,u);
			}
			a.Find_Euler(1);
		}
	}
}

  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-1])这个地方也也有笔误
    应改为L(X [0 .. M-1],Y [0 .. N-1])= 1 + L(X [0 .. M-2],Y [0 .. N-2])