POJ2230Watchcow[欧拉回路]

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 7512		Accepted: 3290		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

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...

Source

USACO 2005 January Silver

---

差点想正反分开求两次，其实只求一遍无向图欧拉回路就可以了

简单说一下构造吧：

对于一个点搜索所有出边最后输出自己

 

//
//  main.cpp
//  poj2230
//
//  Created by Candy on 9/28/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=1e4+,M=5e4+,INF=1e9+;
inline int read(){
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
    return x;
}
int n,m,u,v,w;
struct edge{
    int v,ne;
}e[M<<];
int h[N],cnt=;
inline void ins(int u,int v){
    cnt++;
    e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
    cnt++;
    e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;
}
int vis[M<<];
void dfs(int u){
    for(int i=h[u];i;i=e[i].ne){
        int v=e[i].v;
        if(vis[i]) continue;
        vis[i]=;
        dfs(v);
    }
    printf("%d\n",u);
}
int main(int argc, const char * argv[]) {
    n=read();m=read();
    for(int i=;i<=m;i++){u=read();v=read();ins(u,v);}
    dfs();
    return ;
}