2 seconds
256 megabytes
standard input
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 v, v, ..., v that v = a, v = b and there is a road between v and v 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 u, u, ..., u that u = c, u = d and there is a road between u and u 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 (v, ..., v) and (u, ..., u) 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 v, v, ..., v where v = a and v = b. The second line should contain n distinct integers u, u, ..., u where u = c and u = d.
Two paths generate at most 2n - 2 roads: (v, v), (v, v), ..., (v, v), (u, u), (u, u), ..., (u, u). 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.
题意：看a,b之间，c,d之间是否遍历所有顶点且一次后有路，且路的数目不超过K.
题解：
构造一条a,c……d,b的路
和一条 c,a……b,d的路，最少需要n+1条，所以k>=n+1 且n == 4时，a,b之间和c,d之间都不能有通路，所以n == 4怎么都不行。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
int vis[];
void solve(){
int n,k;
int a,b,c,d;
scanf("%d %d",&n,&k);
scanf("%d%d%d%d",&a,&b,&c,&d);
if(n == || k<n+) printf("-1\n");
else{
vis[a] = vis[b] = vis[c] = vis[d] = ;
printf("%d %d ",a,c);
for(int i = ; i<=n; i++){
if(!vis[i]) printf("%d ",i);
}
printf("%d %d\n",d,b);
printf("%d %d ",c,a);
for(int i = ; i<=n; i++){
if(!vis[i]) printf("%d ",i);
}
printf("%d %d\n",b,d);
}
}
int main()
{
solve();
return ;
}