http://poj.org/problem?id=2230

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...

题解:

一个图是欧拉图,那么他的子图也是一个欧拉图,只需dfs即可

#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1e5+10;
struct node{
int v;
int next;
}G[MAXN];
int head[MAXN],cnt;
bool vis[MAXN];
int ans[MAXN];
void add(int u,int v)
{
G[++cnt].v=v;
G[cnt].next=head[u];
head[u]=cnt;
}
int n,m,k=0;
int dfs(int u)
{
for (int i = head[u]; i!=-1 ; i=G[i].next) {
if(!vis[i])
{
vis[i]= true;
dfs(G[i].v);
ans[k++]=G[i].v;
}
}
}
int main() {
while (scanf("%d%d",&n,&m)!=EOF)
{
cnt=0,k=0;
memset(head,-1, sizeof(head));
memset(vis,false, sizeof(vis));
int u,v;
for (int i = 0; i <m ; ++i) {
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
dfs(1);
for (int i = 0; i <k ; ++i) {
printf("%d\n",ans[i]);
}
printf("1\n");
}
return 0;
}
//poj2230

  

poj2230 欧拉回路的更多相关文章

  1. 0x66 Tarjan算法与无向图联通性

    bzoj1123: [POI2008]BLO poj3694 先e-DCC缩点,此时图就变成了树,树上每一条边都是桥.对于添加边的操作,相当于和树上一条路径构环,导致该路径上所有边都不成为桥.那么找这 ...

  2. Watchcow(POJ2230+双向欧拉回路+打印路径)

    题目链接:http://poj.org/problem?id=2230 题目: 题意:给你m条路径,求一条路径使得从1出发最后回到1,并满足每条路径都恰好被沿着正反两个方向经过一次. 思路:由于可以回 ...

  3. POJ2230 Watchcow【欧拉回路】

    Watchcow Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 6172Accepted: 2663 Special Judge ...

  4. poj2230 Watchcow【欧拉回路】【输出路径】(遍历所有边的两个方向)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4392 题目大意: 一个图,要将每条边恰好遍历两遍,而且要以不同的方向,还要回到原点. dfs解法    ...

  5. POJ2230(打印欧拉回路)

    Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7473   Accepted: 3270   Specia ...

  6. POJ2230Watchcow[欧拉回路]

    Watchcow Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7512   Accepted: 3290   Specia ...

  7. ACM/ICPC 之 混合图的欧拉回路判定-网络流(POJ1637)

    //网络流判定混合图欧拉回路 //通过网络流使得各点的出入度相同则possible,否则impossible //残留网络的权值为可改变方向的次数,即n个双向边则有n次 //Time:157Ms Me ...

  8. [poj2337]求字典序最小欧拉回路

    注意:找出一条欧拉回路,与判定这个图能不能一笔联通...是不同的概念 c++奇怪的编译规则...生不如死啊... string怎么用啊...cincout来救? 可以直接.length()我也是长见识 ...

  9. ACM: FZU 2112 Tickets - 欧拉回路 - 并查集

     FZU 2112 Tickets Time Limit:3000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u P ...

随机推荐

  1. FUNCTION_POWER

    1.power function Definition:The Oracle PL/SQL, the POWER function is a built in function which takes ...

  2. JSP-Servlet中文乱码

    客户端 get 方法时 出现乱码: 解决办法: String str1 = request.getParameter("stuname"); String str = new St ...

  3. June 16th 2017 Week 24th Friday

    Progress is the activity of today and the assurance of tomorrow. 进步是今天的活动,明天的保证. The best preparatio ...

  4. 技术blog

    看到好多人都写技术blog,发现自己工作了快两年居然都没有写过blog,很是惭愧,遂从今天开始不定期更新技术体验,主要是为了记下自己学习摸索技术过程中的经验以及问题解决方案.

  5. Locust性能测试3 no-web运行

    Locust也支持no-web的方式运行,直接通过控制台设置并发用户数.每秒启动用户数.持续压测时间. locust -f 脚本路径 -c 用户数 -r 每秒启动用户数 --run-time 持续压测 ...

  6. MHA 一主一从搭建

    安装基本环境介绍 一主一从     MHA manager & slave  node2 192.168.56.27 Master机器 node1 192.168.56.26 VIP   19 ...

  7. 【洛谷5287】[HNOI2019] JOJO(主席树优化KMP)

    点此看题面 大致题意: 每次往一个字符串末尾加上\(x\)个字符\(c\),或者回到某一历史版本,求\(KMP\)的\(\sum Next_i\). 问题转化 考虑到可以离线. 于是,我们就可以用一个 ...

  8. ACM-ICPC (10/17)

    今天满课啊,天气太冷了,网上找了一下虚树的东西,还没弄懂~~~(:´д`)ゞ 牛客网挑战赛1 给定一棵n个点的树,问其中有多少条长度为偶数的路径.路径的长度为经过的边的条数.x到y与y到x被视为同一条 ...

  9. 2018.12.24 Spring中的aop演示(也就是运用aop技术实现代理模式)

    Aop的最大意义是:在不改变原来代码的前提下,也不对源代码做任何协议接口要求.而实现了类似插件的方式,来修改源代码,给源代码插入新的执行代码. 1.spring中的aop演示 aop:面向方面编程.不 ...

  10. mysql自连接

    自连接: -- 求7369员工编号,姓名,经理编号和经理姓名 select e1.empno,e1.ename,e2.empno,e2.ename from emp e1,emp e2 where e ...