POJ 2230

Watchcow
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 8555 Accepted: 3724 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

题意:
给你一个无向图,每条边要走2次,要你求出点的序列
思路:
因为每条边走2次,而且是反向,所以把它看成有向图,出度一定等于入度,所以就是一道裸的欧拉回路
代码:

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<set>
using namespace std;
struct node {
    int to;
    int num;
    node(int a, int b)
    {
        to = a;
        num = b;
    }
};
vector<node>arr[10100];
int n, m,ans[10100],k=0;
bool book[500100];
void oula(int a)
{
    for (int i = 0; i < arr[a].size(); i++)
    {
        if (!book[arr[a][i].num])
        {
            book[arr[a][i].num] = 1;
            oula(arr[a][i].to);
        }
    }
    cout << a << endl;
}
int main()
{
    cin >> n >> m;
    for (int i = 1; i <= m; i++)
    {
        int a, b;
        cin >> a >> b;
        arr[a].push_back(node(b, i));
        arr[b].push_back(node(a, i+m));
    }
    oula(1);
    return 0;
}