| Time Limit: 3000MS | Memory Limit: 65536K | |||
| Total Submissions: 7512 | Accepted: 3290 | Special Judge | ||
Description
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
* Lines 2..M+1: Two integers denoting a pair of fields connected by a path.
Output
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
Bessie starts at 1 (barn), goes to 2, then 3, etc...
Source
差点想正反分开求两次，其实只求一遍无向图欧拉回路就可以了
//
// main.cpp
// poj2230
//
// Created by Candy on 9/28/16.
// Copyright © 2016 Candy. All rights reserved.
// #include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=1e4+,M=5e4+,INF=1e9+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x;
}
int n,m,u,v,w;
struct edge{
int v,ne;
}e[M<<];
int h[N],cnt=;
inline void ins(int u,int v){
cnt++;
e[cnt].v=v;e[cnt].ne=h[u];h[u]=cnt;
cnt++;
e[cnt].v=u;e[cnt].ne=h[v];h[v]=cnt;
}
int vis[M<<];
void dfs(int u){
for(int i=h[u];i;i=e[i].ne){
int v=e[i].v;
if(vis[i]) continue;
vis[i]=;
dfs(v);
}
printf("%d\n",u);
}
int main(int argc, const char * argv[]) {
n=read();m=read();
for(int i=;i<=m;i++){u=read();v=read();ins(u,v);}
dfs();
return ;
}