[欧拉] poj 2230 Watchcow

时间：2024-09-02 16:34:20

主题链接：

http://poj.org/problem?

id=2230

Watchcow
Time Limit: 3000MS   Memory Limit: 65536K
Total Submissions: 6055   Accepted: 2610   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

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题目意思：

给一幅连通无向图。求如何从点1出发，最后回到1，且每条边两个方向各走一次。

解题思路：

欧拉回路

将一条边抽象成两条有向边就可以。

代码：

//#include<CSpreadSheet.h>

#include<iostream>

#include<cmath>

#include<cstdio>

#include<sstream>

#include<cstdlib>

#include<string>

#include<string.h>

#include<cstring>

#include<algorithm>

#include<vector>

#include<map>

#include<set>

#include<stack>

#include<list>

#include<queue>

#include<ctime>

#include<bitset>

#include<cmath>

#define eps 1e-6

#define INF 0x3f3f3f3f

#define PI acos(-1.0)

#define ll __int64

#define LL long long

#define lson l,m,(rt<<1)

#define rson m+1,r,(rt<<1)|1

#define M 1000000007

//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;

#define Maxn 11000

int n,m;

struct Node

{

    int to,id;

    Node(int a,int b)

    {

        to=a,id=b;

    }

};

vector<vector<Node> >myv;

bool vis[Maxn*10];

vector<int>ans;

void dfs(int cur)

{

    for(int i=0;i<myv[cur].size();i++)

    {

        Node ne=myv[cur][i];

        if(vis[ne.id])

            continue;

        vis[ne.id]=true;

        dfs(ne.to);

    }

    ans.push_back(cur);

}

int main()

{

    //freopen("in.txt","r",stdin);

   //freopen("out.txt","w",stdout);

   while(~scanf("%d%d",&n,&m))

   {

       myv.clear();myv.resize(n+1);

       for(int i=1;i<=m;i++)

       {

           int a,b;

           scanf("%d%d",&a,&b);

           myv[a].push_back(Node(b,i*2-1));

           myv[b].push_back(Node(a,i*2));

       }

       memset(vis,false,sizeof(vis));

       ans.clear();

       dfs(1);

       for(int i=ans.size()-1;i>=0;i--)

            printf("%d\n",ans[i]);

   }

    return 0;

}

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