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

D. Bear and Two Paths

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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 integersu1, 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.

Examples

input

7 11
2 4 7 3

output

2 7 1 3 6 5 4
7 1 5 4 6 2 3

input

1000 999
10 20 30 40

output

-1

Note

In the first sample test, there should be 7 cities and at most 11 roads.
The provided sample solution generates 10 roads, as in the drawing. You can also see a simple path of length n between 2 and 4,
and a path between 7 and 3.



题意：有n个城市，分别是1-n。两个城市间只有间最多只有一条路，城市的总路数不超过k。现给你a，b，c，d。如果满足下列条件

1.a与b之间没有路，存在以a为起点b为终点，把每个城市通过一次的走法。

2.c与d之间没有路，存在以c为起点d为终点，把每个城市通过一次的走法。

则分别件a到b和c到d的走法打印出来（任意一组都行）。

否则则打印-1.

解法：我们可以先考1这种情况，可以设成a-c-x1-x2-x3-x4-x5-........-xn-d-b要经过所有城市，所以这样至少需要有n-1条路（无论c，d在哪都是n-1）。然后对于第2种情况，我们只需要在a到x1少加条路，b到x6加条路即可（若只有x1的话，则需将a与x1，b与x1加条路。n为4的时候无论怎样都不会存在），其输出方案就为c-a-x1-x2-x3-x4-x5-............-x6-b-d。所以k至少需要n-1+2。这样我们在k<n+1||n==4时就输出-1.否则就打应上面的方案。

AC：

#include <stdio.h>

#include <algorithm>

#include<iostream>

#include<map>

#include<string.h>

#include<math.h>

using namespace std;

const int maxn=6005;

int a[maxn];

void swapa(int x,int y)

{

int temp=a[x];

a[x]=a[y];

a[y]=temp;

}

int main()

{

int n,k;

int aa,b,c,d;

scanf("%d%d",&n,&k);

scanf("%d%d%d%d",&aa,&b,&c,&d);

for(int i=0;i<=n;i++)

{

a[i]=i;

}

if(k<n+1||n==4) printf("-1\n") ;

else

{

for(int i=1;i<=n;i++)

{

if(a[i]==aa) swapa(i,1);

}

for(int i=1;i<=n;i++)

{

if(a[i]==b) swapa(i,n);

}

for(int i=1;i<=n;i++)

{

if(a[i]==c) swapa(i,2);

}

for(int i=1;i<=n;i++)

{

if(a[i]==d) swapa(i,n-1);

}

for(int i=1;i<=n;i++)

{

printf("%d ",a[i]);

}

printf("\n");

swapa(1,2);

swapa(n-1,n);

for(int i=1;i<=n;i++)

{

printf("%d ",a[i]);

}

printf("\n");

}

}