题目地址：POJ 2230
最普通的欧拉回路。边不重复记录点。不多说。
代码如下：
#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>#include <map>#include <set>#include <stdio.h>using namespace std;#define LL long long#define pi acos(-1.0)const int mod=1e9+7;const int INF=0x3f3f3f3f;const double eqs=1e-9;int head[11000], cnt, vis[120000], path[120000], tot;struct node{int u, v, id, next;}edge[120000];void add(int u, int v, int id){edge[cnt].v=v;edge[cnt].id=id;edge[cnt].next=head[u];head[u]=cnt++;}void dfs(int u){for(int i=head[u];i!=-1;i=edge[i].next){if(!vis[edge[i].id]){vis[edge[i].id]=1;head[u]=i;dfs(edge[i].v);}i=head[u];}path[tot++]=u;}void init(){memset(head,-1,sizeof(head));cnt=tot=0;memset(vis,0,sizeof(vis));}int main(){int n, m, i, u, v;scanf("%d%d",&n,&m);init();for(i=0;i<m;i++){scanf("%d%d",&u,&v);add(u,v,i<<1);add(v,u,i<<1|1);}dfs(1);for(i=tot-1;i>=0;i–){printf("%d\n",path[i]);}return 0;}
，怠惰是贫穷的制造厂。
