D. Bear and Two Paths

题目连接:

http://www.codeforces.com/contest/673/problem/D

Description

Bearland has n cities, numbered 1 through n. Cities are connected via bidirectional roads. Each road connects two distinct cities. No two roads connect the same pair of cities.

Bear Limak was once in a city a and he wanted to go to a city b. There was no direct connection so he decided to take a long walk, visiting each city exactly once. Formally:

There is no road between a and b.

There exists a sequence (path) of n distinct cities v1, v2, ..., vn that v1 = a, vn = b and there is a road between vi and vi + 1 for .

On the other day, the similar thing happened. Limak wanted to travel between a city c and a city d. There is no road between them but there exists a sequence of n distinct cities u1, u2, ..., un that u1 = c, un = d and there is a road between ui and ui + 1 for .

Also, Limak thinks that there are at most k roads in Bearland. He wonders whether he remembers everything correctly.

Given n, k and four distinct cities a, b, c, d, can you find possible paths (v1, ..., vn) and (u1, ..., un) to satisfy all the given conditions? Find any solution or print -1 if it's impossible.

Input

The first line of the input contains two integers n and k (4 ≤ n ≤ 1000, n - 1 ≤ k ≤ 2n - 2) — the number of cities and the maximum allowed number of roads, respectively.

The second line contains four distinct integers a, b, c and d (1 ≤ a, b, c, d ≤ n).

Output

Print -1 if it's impossible to satisfy all the given conditions. Otherwise, print two lines with paths descriptions. The first of these two lines should contain n distinct integers v1, v2, ..., vn where v1 = a and vn = b. The second line should contain n distinct integers u1, u2, ..., un where u1 = c and un = d.

Two paths generate at most 2n - 2 roads: (v1, v2), (v2, v3), ..., (vn - 1, vn), (u1, u2), (u2, u3), ..., (un - 1, un). Your answer will be considered wrong if contains more than k distinct roads or any other condition breaks. Note that (x, y) and (y, x) are the same road.

Sample Input

7 11

2 4 7 3

Sample Output

2 7 1 3 6 5 4

7 1 5 4 6 2 3

题意

给你n和k,表示这个图有n个点,最多k条边

现在让你构造这个图,使得存在两条路径a-v2-...-b,c-v2-...-d恰好都经过n个点

且ab不相邻,cd不相邻

题解:

很显然构造一个蝴蝶的样子就好了,左边一个三角形,右边一个三角形,中间是一条链

这样需要n+1条边就可以了

由于ab不能挨在一起,所以4的时候,是不可行的

代码

#include<bits/stdc++.h>
using namespace std;
int n,k,a,b,c,d;
vector<int>tmp;
int main()
{ scanf("%d%d",&n,&k);
scanf("%d%d%d%d",&a,&b,&c,&d);
if(n==4)return puts("-1"),0;
if(k<=n)return puts("-1"),0;
for(int i=1;i<=n;i++)
{
if(i==a||i==b||i==c||i==d)continue;
tmp.push_back(i);
}
cout<<a<<" "<<c<<" ";
for(int i=0;i<tmp.size();i++)cout<<tmp[i]<<" ";
cout<<d<<" "<<b<<endl;
cout<<c<<" "<<a<<" ";
for(int i=0;i<tmp.size();i++)cout<<tmp[i]<<" ";
cout<<b<<" "<<d<<endl;
}

Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D. Bear and Two Paths 构造的更多相关文章

  1. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) B. Problems for Round 水题

    B. Problems for Round 题目连接: http://www.codeforces.com/contest/673/problem/B Description There are n ...

  2. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) B

    B. Problems for Round time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

  3. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition)只有A题和B题

    连接在这里,->点击<- A. Bear and Game time limit per test 2 seconds memory limit per test 256 megabyte ...

  4. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D Bear and Two Paths

    题目链接: http://codeforces.com/contest/673/problem/D 题意: 给四个不同点a,b,c,d,求是否能构造出两条哈密顿通路,一条a到b,一条c到d. 题解: ...

  5. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C - Bear and Colors

    题目链接: http://codeforces.com/contest/673/problem/C 题解: 枚举所有的区间,维护一下每种颜色出现的次数,记录一下出现最多且最小的就可以了. 暴力n*n. ...

  6. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) C. Bear and Colors 暴力

    C. Bear and Colors 题目连接: http://www.codeforces.com/contest/673/problem/C Description Bear Limak has ...

  7. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) A. Bear and Game 水题

    A. Bear and Game 题目连接: http://www.codeforces.com/contest/673/problem/A Description Bear Limak likes ...

  8. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition)

    A.暴力枚举,注意游戏最长为90分钟 B.暴力,c[l]++,c[r]--,记录中间有多长的段是大小为n的,注意特判m=0的情况 C.暴力枚举,我居然一开始没想出来!我一直以为每次都要统计最大的,就要 ...

  9. Codeforces Round #351 (VK Cup 2016 Round 3, Div. 2 Edition) D

    D. Bear and Two Paths time limit per test 2 seconds memory limit per test 256 megabytes input standa ...

随机推荐

  1. RW RO ZI ROM keil中的含义

    编译的一个ARM的程序,会得到这样的信息: ============================================================================== ...

  2. 88.modelsim仿真do文件相关技巧

    网上的关于DO文件的编写好像资料不多,比较杂,所以本人总结一下常用的简单语法,方便大家查看.其实本人也刚接触DO文件没多久,有纰漏很正常,欢迎指正批评,互相学习.PS:写得有点乱   还有一个值得注意 ...

  3. 如何禁止Linux内核的-O2编译选项【转】

    转自:http://blog.csdn.net/larryliuqing/article/details/8674274 http://lenky.info/2013/03/10/%E5%A6%82% ...

  4. dstat 服务器性能查看命令【转】

    一. 安装和简解 # yum -y install dstat# dstat CPU状态:CPU的使用率.这项报告更有趣的部分是显示了用户,系统和空闲部分,这更好地分析了CPU当前的使用状况.如果你看 ...

  5. [HBase]region split流程

    1. 简介 HBase 的最小管理单位为region,region会按照region 分裂策略进行分裂. 基于CDH5.4.2 2. 总览

  6. spring boot 中使用redis session

    spring boot 默认的httpsession是存在内存中.这种默认方式有几个缺点:1.当分布式部署时,存在session不一致的问题:2.当服务重启时session就会丢失,这时候用户就需要重 ...

  7. 列表CListCtrl类使用

    CListCtrl是列表控件类,列表控件的每一行叫做一个item,每一列叫做一个subitem.每一行和每一列都有个ID号,可以确定唯一的单元格. 最近使用了这个控件,有心得总结如下: (Dialog ...

  8. Linux命令之cp命令

    cp命令:用来将一个或多个源文件或者目录复制到指定的目的文件或目录.它可以将单个源文件复制成一个指定文件名的具体的文件或一个已经存在的目录下.cp命令还支持同时复制多个文件,当一次复制多个文件时,目标 ...

  9. Rookey.Frame之实体FluentValidation验证

    昨天给大家介绍了Rookey.Frame框架的实体设计,今天继续跟大家分享实体的FluentValidation验证,在Rookey.Frame框架中可以设置多种验证方式:FluentValidati ...

  10. java.lang.NoClassDefFoundError: ognl/PropertyAccessor

    本篇对 Web 开发中,项目部署后.开启 Tomcat 服务器 Console 控制台报错 java.lang.NoClassDefFoundError: ognl/PropertyAccessor ...