POJ2230 Watchcow(欧拉回路)

最新推荐文章于 2019-07-31 11:03:51 发布

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#欧拉回路

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本文介绍了一种算法解决方案，帮助奶牛Bessie在农场中构建一条特殊的巡逻路径，确保她能按照要求走过每条路径两次且方向相反。通过构造图论中的欧拉回路，文章提供了一个具体的实现案例。

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6467		Accepted: 2819		Special Judge

Description

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.

Input

* Line 1: Two integers, N and M.

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

Output

* 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.

Sample Input

Sample Output

Hint

OUTPUT DETAILS:

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

题目大意:

欧拉回路的题目，要求来回各走一遍,也就是每条边要遍历两次，故肯定存在欧拉回路.

AC代码:

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn = 50005;
int head[maxn<<1];  //代表结点 
bool vis[maxn<<1];

struct EdgeNode
{
	int to;
	int w;
	int next;	
}Edges[maxn*3];

int k;

void init()
{
	memset(vis,false,sizeof(vis));
	memset(head,-1,sizeof(head));
	k = 0;	
}
void add_Edge(int u,int v)
{
	//Edges[k].w = w;
	Edges[k].to = v;
	Edges[k].next = head[u];  //head[u]代表一个结点 
	head[u] = k++;
}

void dfs(int x)
{
	int i;
	for(i=head[x];i!=-1;i=Edges[i].next) //遍历父结点为x的所有边 
	{
		if(!vis[i])
		{
			vis[i] = true;
			dfs(Edges[i].to);
		}
	}
	printf("%d\n",x);
}

int main()
{
	int n,m;
	int num1,num2;
	//freopen("1.txt","r",stdin);
	while(scanf("%d%d",&n,&m) != EOF)
	{
		int i;
		init();
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&num1,&num2);
			add_Edge(num1,num2);
			add_Edge(num2,num1);
		}
		dfs(1);
	}
	return 0;
}