POJ - 2230 欧拉回路的输出

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

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

Sample Output

1 2 3 4 2 1 4 3 2 4 1

Hint

OUTPUT DETAILS: 

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

题意：

现在有N个农场和M条路,要求你走过这M条路每条2次且这两次是反向的.要你输出从1号农场走到完所有M条路两次之后回到1号农场的点的轨迹.保证这种轨迹存在.其中点(农场)从1开始编号到N.

分析：

因为题目保证存路径，所以直接dfs就行，标记边是否走过即可，因为有重边，所以用链式前向星保存。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#define ll long long
#define PI 3.1415926535897932384626
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=100050;
struct Edge
{
    int to,next,id;
    bool flag;
}edge[maxn*10];
int head[maxn];
int in[maxn];
int tot;
 
void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
   // edge[tot].id=w;
    edge[tot].flag=false;
    head[u]=tot++;
}
int start;
vector<int>ans;
void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
    memset(in,0,sizeof(in));
}
void dfs(int x)
{

    for(int i=head[x];i!=-1;i=edge[i].next)
    {
        if(!edge[i].flag)
        {
            edge[i].flag=true;
           // edge[i^1].flag=true;
            dfs(edge[i].to);
           // ans.push_back(i);//记录边的号码
        }
    }
      cout<<x<<endl;
}
int main()
{
        init();
        int n,m;
        scanf("%d%d",&n,&m);
        int x,y;
        for(int i=1;i<=m;i++)
        {
         scanf("%d%d",&x,&y);
            addedge(x,y);
            addedge(y,x);
            in[x]++;
            in[y]++;
        }
        ans.clear();
        dfs(1);
        
 

    return 0;
}