题干：
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 ndistinct 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 ndistinct 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 kdistinct 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.
题目大意：
给你一个包含N个点的无向图，要求我们用小于等于K条边来构造出这个图。
这个图有两对起点和终点：第一对的起点a，终点b。第二对的起点c，终点d；
现在要求构造出的图，a和b之间没有直接相连的边，c和d之间没有直接相连的边。
但是从a到b有一条路径，a作为起点，b作为终点，路径上包含所有点（1.2.3.4.5..............N）。
同理，从c到d也有一条路径，c作为起点，d作为终点，路径上也包含所有点。
如果可以构造出来，输出这两条路径经过的点；反之输出-1。
解题报告：
因为是构造题，并且要求能否在K条边之内构造出来，所以肯定要构造一个最少边数的解。首先a为起点b为终点的肯定需要n-1条边，c到d还需要连边，所以肯定最少需要n条边。但是n条边肯定无法使得ac都可以当起点，bd当终点。所以需要n+1条边，发现可以构造，且总可以构造，也就是a->c->a->...->b->d->b的一条路径，即可满足条件，并且第二个a和b中间必须要有点，这也就要求点数必须>5，否则不满足ab不相邻的条件，于是这样构造即可。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define FF first
#define SS second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,k,a,b,c,d;
int main()
{cin>>n>>k;cin>>a>>b>>c>>d;if(k < n+1 || n == 4) {printf("-1\n");return 0;} printf("%d %d ",a,c);for(int i = 1; i<=n; i++) {if(i == a || i == b || i == c || i == d) continue;printf("%d ",i); }printf("%d %d\n",d,b);printf("%d %d ",c,a);for(int i = 1; i<=n; i++) {if(i == a || i == b || i == c || i == d) continue;printf("%d ",i); }printf("%d %d\n",b,d);return 0 ;
}
