(with Tuola loop solution) POJ 2230

Watchcow

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 6255		Accepted: 2708		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 is doing any evil. She begins at the barn, makes his patrol, and then returns to the barn when she's done.

If She were a more observant cow, she might being able to just walk each of M (1 <= M <= 50,000) bidirectional trails N Umbered 1..M between N (2 <= n <=) fields numbered 1..N on the farm once and is confident that she ' s seen Eve Rything she needs to see. But since she is ' t, she wants to make sure she walks down each trail exactly twice. It's also important that she's the trips along each trail is in opposite directions, so, she doesn ' t miss the same thing Twice.

A pair of fields might is connected by more than one trail. Find a path that Bessie can follow which would 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 are possible, output any solution.

Sample Input

4 51 21 42 32 43 4

Sample Output

12342143241

Hint

OUTPUT DETAILS:

Bessie starts at 1 (barn), goes to 2, then 3, etc ...

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath > #include <cstdlib> #include <algorithm> #include <queue> #include <vector> #include < Stack>using namespace Std;int n,m;vector<int> e[10010];bool vis[10010];void dfs (int u) {for    (int i=0;i< E[u].size (); i++)    {        int v=e[u][i];        if (v!=-1)        {            e[u][i]=-1;            DFS (v);        }    }    printf ("%d\n", U);} int main () {    int x, y;    memset (vis,0,sizeof (Vis));    scanf ("%d%d", &n,&m);    for (int i=0;i<m;i++)    {        scanf ("%d%d", &x,&y);        E[x].push_back (y);        E[y].push_back (x);    }    DFS (1);    return 0;}

　　

(with Tuola loop solution) POJ 2230