【poj2230】Watchcow

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

题解

给定n个点m条边的无向图G=(V,E)，求一回路，经过所有边两次且仅两次，且要求第二次经过同一条边时行进的方向与第一次不同。

先把一条边拆成两条边，转化成有向图，然后欧拉回路。

#include<iostream>
#include<cstdio>
#define M 50010
#define N 10010
using namespace std;
struct edge_node
{
    int to,next;
    bool vis;
}e[M*2];
int head[N];
int cnt;
void ins(int u,int v)
{
    e[++cnt].to = v; e[cnt].next = head[u]; head[u] = cnt;
    e[cnt].vis = false;
}
void find(int u)
{
    for (int i=head[u];i;i=e[i].next)
    {
        if (!e[i].vis)
        {
            e[i].vis = true;
            find(e[i].to);
        }
    }
    printf("%d
",u);
}
int main()
{
    int n,m,u,v;
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d",&u,&v);
        ins(u,v);
        ins(v,u);
    }
    find(1);
}