就是欧拉判定，判定之后就可以使用DFS求欧拉回路了。图论内容。

这里使用邻接矩阵会快很多速度。

这类题目都是十分困难的，光是定义的记录的数组变量就会是一大堆。

#include <cstdio>
#include <cstring>
#include <stack>
#include <vector>
using namespace std;

struct Edge
{
	int ed, des;
	Edge(int e = 0, int d = 0) : ed(e), des(d) {}
};
const int EDGES = 2000;//1996;
const int VEC = 45;
stack<int> stk;
int degree[VEC];
vector<Edge> gra[VEC];
bool vis[EDGES];

void euler(int u)
{
	for(int i = 0; i < (int)gra[u].size(); i++)
	{
		if(!vis[gra[u][i].ed])	//标志访问过了,这里需要表示桥，不是顶点
		{
			vis[gra[u][i].ed] = true;
			euler(gra[u][i].des);
			stk.push(gra[u][i].ed);
			//不能break
		}
	}
}

int main()
{
	int x, y, z, one;	
	while (scanf("%d %d", &x, &y) != EOF && x && y)
	{
		memset(vis, 0, sizeof(vis));
		memset(degree, 0, sizeof(degree));
		for (int i = 1; i < VEC; i++)
			gra[i].clear();

		scanf("%d", &z);
		one = min(x, y);
		degree[x]++; degree[y]++;
		gra[x].push_back(Edge(z, y)); gra[y].push_back(Edge(z, x));
		while (scanf("%d %d", &x, &y) != EOF && x && y)
		{
			scanf("%d", &z);
			gra[x].push_back(Edge(z, y)); gra[y].push_back(Edge(z, x));
			degree[x]++, degree[y]++;
		}
		for (int i = 1; i < VEC; i++)
		{
			if (degree[i] & 1)
			{
				puts("Round trip does not exist.");
				goto endLoop;	//玩玩goto
			}
		}
		
		euler(one);
		while (!stk.empty())
		{
			printf("%d ", stk.top());
			stk.pop();
		}
		putchar('\n');

		endLoop:;
	}
	return 0;
}