POJ2230 Watchcow

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8431		Accepted: 3666		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

[b]【题解】[/b]
欧拉路水题，顺便复习一波欧拉路 http://www.cnblogs.com/nextbin/p/4005396.html 这个博客不错

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

inline void read(int &x)
{
x = 0;char ch = getchar(), c = ch;
while(ch < '0' || ch > '9')c = ch, ch = getchar();
while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
if(c == '-')x = -x;
}

const int INF = 0x3f3f3f3f;
const int MAXM = 50000 + 10;
const int MAXN = 100000 + 10;

struct Edge
{
int u,v,next;
Edge(int _u, int _v, int _next){u = _u;v = _v;next = _next;}
Edge(){}
}edge[MAXM << 1];
int head[MAXN], cnt;
inline void insert(int a, int b)
{
edge[++cnt] = Edge(a,b,head[a]);
head[a] = cnt;
}
int n,m;
int b[MAXM << 1];
int dfs(int u)
{
for(register int pos = head[u];pos;pos = edge[pos].next)
{
if(b[pos]) continue;
b[pos] = 1;
dfs(edge[pos].v);
}
printf("%d\n", u);
}

int main()
{
read(n), read(m);
int tmp1,tmp2;
for(register int i = 1;i <= m;++ i)
read(tmp1), read(tmp2), insert(tmp1, tmp2), insert(tmp2, tmp1);
dfs(1);
return 0;
}

POJ2230