Poj 1041 John's trip (Fleury算法求欧拉回路路径)

题意：给出无向图，每条边有唯一的序号，是否存在欧拉回路，若存在输出边序号最小字典序的路径。

思路：Fleury算法求欧拉回路路径，Fleury算法其实就是DFS套了个人名……这里有个介绍http://blog.csdn.net/niushuai666/article/details/6549182

依据静态邻接表的特性，从序号大到小建立边，这样即可保证序号最小的边首先访问到

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define upmax(a,b) ((a)=(a)>(b)?(a):(b))

const int N=2010;
const int INF=0x3fffffff;

int n,e,head
;
bool visit
;
int degree
,id,answer
;

struct Node
{
	int v,way,pd,next;
}edge[N<<3];

struct Data
{
	int a,b,c;
	bool operator < (const Data b) const
	{
		return c<b.c;
	}
}data[N<<3];

void Add (int a,int b,int c)
{
	edge[e].v=b;
	edge[e].way=c;
	edge[e].next=head[a];
	head[a]=e++;
}

void Fleury (int start)
{
	for (int i=head[start];i!=-1;i=edge[i].next)
	{
		int v=edge[i].v;
		if (visit[edge[i].way]==false)
		{
			visit[edge[i].way]=true;
			Fleury(v);
			answer[++id]=edge[i].way;  //记录边的访问顺序,由于回溯,所以id越大越先访问
		}
	}
}

int main ()
{
	int i,limit=0,x,y,z;
	while (scanf("%d%d",&x,&y),x||y)
	{
		int cnt=0;
		e=id=0;
		memset(visit,false,sizeof(visit));
		memset(data,0,sizeof(data));
		memset(head,-1,sizeof(head));
		memset(degree,0,sizeof(degree));

		scanf("%d",&z);
		data[cnt].a=x,data[cnt].b=y,data[cnt++].c=z;
		degree[x]++;
		degree[y]++;
		upmax(limit,x);
		upmax(limit,y);
		while (scanf("%d%d",&x,&y),x||y)
		{
			scanf("%d",&z);
			upmax(limit,x);
			upmax(limit,y);
			degree[x]++;
			degree[y]++;
			data[cnt].a=x,data[cnt].b=y,data[cnt++].c=z;
		}
		bool flag=false;
		for (i=1;i<=limit;i++)
			if (degree[i]%2==1)   //有奇数度点则不存在欧拉回路
			{
				flag=true;
				break;
			}
		if (flag)
		{
			printf("Round trip does not exist.\n");
			continue;
		}
		sort(data,data+cnt);
		for (i=cnt-1;i>=0;i--)
		{
			Add(data[i].a,data[i].b,data[i].c);
			Add(data[i].b,data[i].a,data[i].c);
		}
		Fleury(1);
		for (i=id;i>=1;i--)
			printf(i==1?"%d\n":"%d ",answer[i]);
	}
	return 0;
}