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
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.
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.
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).
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.
7 11 2 4 7 3
2 7 1 3 6 5 4 7 1 5 4 6 2 3
1000 999 10 20 30 40
-1
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.
題意：有n個城市，分別是1-n。兩個城市間只有間最多隻有一條路，城市的總路數不超過k。現給你a，b，c，d。如果滿足下列條件
1.a與b之間沒有路，存在以a爲起點b爲終點，把每個城市通過一次的走法。
2.c與d之間沒有路，存在以c爲起點d爲終點，把每個城市通過一次的走法。
則分別件a到b和c到d的走法打印出來（任意一組都行）。
否則則打印-1.
解法：我們可以先考1這種情況，可以設成a-c-x1-x2-x3-x4-x5-........-xn-d-b要經過所有城市，所以這樣至少需要有n-1條路（無論c，d在哪都是n-1）。然後對於第2種情況，我們只需要在a到x1少加條路，b到x6加條路即可（若只有x1的話，則需將a與x1，b與x1加條路。n爲4的時候無論怎樣都不會存在），其輸出方案就爲c-a-x1-x2-x3-x4-x5-............-x6-b-d。所以k至少需要n-1+2。這樣我們在k<n+1||n==4時就輸出-1.否則就打應上面的方案。
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");
}
}