Problem Statement

You are given a simple connected undirected graph with $N$ vertices and $M$ edges. (A graph is said to be simple if it has no multi-edges and no self-loops.)

For $i = 1, 2, \ldots, M$, the $i$-th edge connects Vertex $u_i$ and Vertex $v_i$.

A sequence $(A_1, A_2, \ldots, A_k)$ is said to be a path of length $k$ if both of the following two conditions are satisfied:

  • For all $i = 1, 2, \dots, k$, it holds that $1 \leq A_i \leq N$.
  • For all $i = 1, 2, \ldots, k-1$, Vertex $A_i$ and Vertex $A_{i+1}$ are directly connected with an edge.

An empty sequence is regarded as a path of length $0$.

You are given a sting $S = s_1s_2\ldots s_N$ of length $N$ consisting of $0$ and $1$.
A path $A = (A_1, A_2, \ldots, A_k)$ is said to be a good path with respect to $S$ if the following conditions are satisfied:

  • For all $i = 1, 2, \ldots, N$, it holds that:
    • if $s_i = 0$, then $A$ has even number of $i$'s.
    • if $s_i = 1$, then $A$ has odd number of $i$'s.

Under the Constraints of this problem, it can be proved that there is at least one good path with respect to $S$ of length at most $4N$.
Print a good path with respect to $S$ of length at most $4N$.

Constraints

  • $2 \leq N \leq 10^5$
  • $N-1 \leq M \leq \min\lbrace 2 \times 10^5, \frac{N(N-1)}{2}\rbrace$
  • $1 \leq u_i, v_i \leq N$
  • The given graph is simple and connected.
  • $N, M, u_i$, and $v_i$ are integers.
  • $S$ is a string of length $N$ consisting of $0$ and $1$.

Input

Input is given from Standard Input in the following format:

$N$ $M$
$u_1$ $v_1$
$u_2$ $v_2$
$\vdots$
$u_M$ $v_M$
$S$

Output

Print a good path with respect to $S$ of length at most $4N$ in the following format.
Specifically, the first line should contain the length $K$ of the path, and the second line should contain the elements of the path, with spaces in between.

$K$
$A_1$ $A_2$ $\ldots$ $A_K$

Sample Input 1

6 6
6 3
2 5
4 2
1 3
6 5
3 2
110001

Sample Output 1

9
2 5 6 5 6 3 1 3 6

The path $(2, 5, 6, 5, 6, 3, 1, 3, 6)$ has a length no greater than $4N$, and

  • it has odd number ($1$) of $1$
  • it has odd number ($1$) of $2$
  • it has even number ($2$) of $3$
  • it has even number ($0$) of $4$
  • it has even number ($2$) of $5$
  • it has odd number ($3$) of $6$

so it is a good path with respect to $S = 110001$.


Sample Input 2

3 3
3 1
3 2
1 2
000

Sample Output 2

0

An empty path $()$ is a good path with respect to $S = 000000$.
Alternatively, paths like $(1, 2, 3, 1, 2, 3)$ are also accepted.

真的要图吗?可以尝试只经过图里面的一棵生成树

考虑数中序列节点相邻的在序列相邻是什么东西?欧拉环游序!

但是欧拉环游序的奇偶不一定正确,怎么办?

想一下如何改变一个位置的奇偶。可以先向他父亲走一步,然后走回来,然后再向父亲走,好像就满足了。

但是 1 没有父亲?

反正一开始都搜到了,在最后回去的时候,特判一下,不走就行了

#include<cstdio>
const int N=1e5+5;
int n,m,k,idx,a[N<<2],b[N<<2],f[N],t[N],fa[N],hd[N],e_num,u,v,c[N];
char s[N];
struct edge{
int v,nxt;
}e[N<<2];
void add_edge(int u,int v)
{
e[++e_num]=(edge){v,hd[u]};
hd[u]=e_num;
}
int find(int x)
{
if(fa[x]==x)
return x;
return fa[x]=find(fa[x]);
}
void dfs(int x,int y)
{
a[++idx]=x;
for(int i=hd[x];i;i=e[i].nxt)
{
if(e[i].v!=y)
{
dfs(e[i].v,x);
a[++idx]=x;
}
}
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
fa[i]=i;
for(int i=1;i<=m;i++)
{
scanf("%d%d",&u,&v);
if(find(u)!=find(v))
fa[find(u)]=find(v),add_edge(u,v),add_edge(v,u);
}
scanf("%s",s+1);
dfs(1,0);
for(int i=idx;i>=1;i--)
if(!t[a[i]])
t[a[i]]=1,f[i]=1;;
for(int i=1;i<idx;i++)
{
if(f[i])
if((c[a[i]]&1)==(s[a[i]]-'0'))
b[++k]=a[i],b[++k]=a[i+1],c[a[i+1]]++,c[a[i]]++;
b[++k]=a[i];
c[a[i]]++;
}
if((c[1]&1)!=(s[1]-'0'))
b[++k]=1;
printf("%d\n",k);
for(int i=1;i<=k;i++)
printf("%d ",b[i]);
}

[ABC244G] Construct Good Path的更多相关文章

  1. Path Creation and Path Painting

    [Path Creation and Path Painting] Path creation and path painting are separate tasks. First you crea ...

  2. imutils.path

    from imutils import paths # 要在哪条路径下查找 path = '...' # 查找图片,得到图片路径 imagePaths = list(imutils.paths.lis ...

  3. alfresco category searches...

    From page 475 of the Alfresco Developer Guide- Category searches use the PATH field, but you constru ...

  4. ArrowLayer : A coustom layer animation

    Since my other answer (animating two levels of masks) has some graphics glitches, I decided to try r ...

  5. Java fundamentals of basic IO

    IO is a problem difficult to handle in various of systems because it  always becomes a bottleneck in ...

  6. react native mac install

    Mac上使用react native tips: 1. 安装Homebrew ruby -e "$(curl -fsSL https://raw.githubusercontent.com/ ...

  7. boost操作xml 5分钟官方教程

    Five Minute Tutorial This tutorial uses XML. Note that the library is not specifically bound to XML, ...

  8. Device trees, Overlays and Parameters of Raspberry Pi

    Raspberry Pi's latest kernels and firmware, including Raspbian and NOOBS releases, now by default us ...

  9. hdfs api读写文写件个人练习

    看下hdfs的读写原理,主要是打开FileSystem,获得InputStream or OutputStream: 那么主要用到的FileSystem类是一个实现了文件系统的抽象类,继承来自org. ...

  10. Android測试环境变量配置表

    要改动的文件是~/bash_profile这个配置文件,内容例如以下: # Setting PATH for Java JAVA_HOME="/Library/Java/JavaVirtua ...

随机推荐

  1. 如何通过API接口获取微店的商品详情

    微店是一款电商平台,对于商家而言,了解商品详情数据是非常重要的.通过API接口获取微店的商品详情,可以让商家更加便捷地管理和分析商品数据.下面就让我们详细了解一下如何通过API获取微店的商品详情. 第 ...

  2. 从驾考科目二到自动驾驶,聊聊GPU为什么对自动驾驶很重要

    "下一个项目,坡道起步." -- "考试不合格,请将车子开到起点,重新验证考试.你的扣分项是:起步时间超30秒:扣100分.行驶过程中车轮轧到边线:扣100分." ...

  3. windows下flutter的环境安装

    Flutter是谷歌出品的移动应用SDK,性能卓越.体验精美.跨平台.HotReload等等这些特点. Dart是谷歌推出的编程语言.支持即时编译JIT(Just In Time).HotReload ...

  4. Spring中事务的传播行为有哪些?

    Spring中事务的传播行为有哪些? 现在我们来谈一个场景,再来引出事务传播行为这个概念.现在有methodA( ) 和 methodB( ),而且两个方法都显示的开启了事务,那么methodB( ) ...

  5. KRPano多屏互动原理

    KRPano可以实现多个屏幕之间的同步显示,主要应用到Websocket技术进行通信. 在控制端,我们需要发送当前KRPano场景的实时的视角和场景信息,可以使用如下的代码: embedpano({ ...

  6. WEB组态编辑器插件(BY组态)介绍

    BY组态是一款非常优秀的纯前端的[web组态插件工具],采用标准HTML5技术,基于B/S架构进行开发,支持WEB端呈现,支持在浏览器端完成便捷的人机交互,简单的拖拽即可完成可视化页面的设计.可无缝嵌 ...

  7. [HUBUCTF 2022 新生赛]ezPython

    附件链接:https://wwvc.lanzouj.com/iIqq218z5x0d 给了一个pyc文件 利用命令将pyc转换为py文件 uncompyle6 ezPython.pyc > ez ...

  8. 01.前后端分离中台框架后端 Admin.Core 学习-介绍与配置说明

    中台框架后端项目 Admin.Core 的介绍与配置说明 中台admin是前后端分离权限管理系统,Admin.Core为后端项目,基于.NET 7.0开发. 支持多租户.数据权限.动态 Api.任务调 ...

  9. 国庆微信头像DIY:轻松打造个性化头像

    前言 国庆节马上要到了,今天就教你如何从0到1使用canvas生成国庆风微信头像. 本文包含以下内容: vue3项目搭建,需求分析 canvas合成图片原理 github自动化部署 开发过程遇到的问题 ...

  10. C中code关键字

    单片机C语言code是什么作用? code的作用是告诉单片机,我定义的数据要存储在ROM(程序存储区)里面,写入后就不能再更改,其实是相当与汇编里面的寻址MOVC(好像是),因为C语言中没办法详细描述 ...