NP-Hard Problem（cf#360）

C. NP-Hard Problem

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Recently, Pari and Arya did some research about NP-Hard problems and they found the minimum vertex cover problem very interesting.

Suppose the graph G is given. Subset A of
its vertices is called a vertex cover of this graph, if for each edge uv there is at
least one endpoint of it in this set, i.e. 

 or 

 (or
both).

Pari and Arya have won a great undirected graph as an award in a team contest. Now they have to split it in two parts, but both of them want their parts of the graph to be a vertex cover.

They have agreed to give you their graph and you need to find two disjoint subsets of its vertices A and B,
such that both A and B are vertex cover
or claim it's impossible. Each vertex should be given to no more than one of the friends (or you can even keep it for yourself).

Input

The first line of the input contains two integers n and m (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000) —
the number of vertices and the number of edges in the prize graph, respectively.

Each of the next m lines contains a pair of integers ui and vi (1  ≤  ui,  vi  ≤  n),
denoting an undirected edge between ui and vi.
It's guaranteed the graph won't contain any self-loops or multiple edges.

Output

If it's impossible to split the graph between Pari and Arya as they expect, print "-1" (without quotes).

If there are two disjoint sets of vertices, such that both sets are vertex cover, print their descriptions. Each description must contain two lines. The first line contains a single integer k denoting
the number of vertices in that vertex cover, and the second line contains kintegers — the indices of vertices. Note that because of m ≥ 1,
vertex cover cannot be empty.

Examples

input

4 2
1 2
2 3

output

1
2
2
1 3

input

3 3
1 2
2 3
1 3

output

-1

Note

In the first sample, you can give the vertex number 2 to Arya and vertices numbered 1 and 3 to
Pari and keep vertex number 4 for yourself (or give it someone, if you wish).

In the second sample, there is no way to satisfy both Pari and Arya.

题意是要将一个图的顶点分成两个互不相交的子集，也就是说每一条边的两个顶点不能同时给一个人，需要把每条边都切开。

给每个顶点都构建一个邻接表，把该点先染为一种色，再把他所有的邻接点都染为另一个色，跑完一轮bfs。每个点都需要先判断一次是否已经被染色，若没有则对其进行一轮染色，以此保证不是联通图也能把每个点都进行了染色。

#include <iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
const int M=1e5+5;
int h[M],cnt=0,c[M];

struct node
{
int next,to;
}e[M*2];
void add(int u,int v)            //构建邻接表
{
e[cnt].to=v;
e[cnt].next=h[u];
h[u]=cnt++;
}
bool bfs(int s)
{
c[s]=1;
queue<int>q;
q.push(s);
while(!q.empty())
{
int x=q.front();
q.pop();
for(int i=h[x];i+1;i=e[i].next)                //对该点的所有邻接点染色
{
int t=e[i].to;
if(c[t]==-1)
{
c[t]=!c[x];
q.push(t);
}
else if(c[x]==c[t]) return 1;
}
}
return 0;
}
int main()
{
int n,m,u,v;
memset(h,-1,sizeof(h));
memset(c,-1,sizeof(c));
scanf("%d%d",&n,&m);
for(int i=0;i<m;i++)
{
scanf("%d%d",&u,&v);
add(u,v);
add(v,u);
}
for(int i=1;i<=n;i++)                //对每个点判断是否被染色
{
if(c[i]==-1)
{
if(bfs(i))
{
puts("-1");
return 0;
}
}
}
int aa=0,bb=0;
int b[M],a[M];
for(int i=1;i<=n;i++)
{
if(c[i]==1)
b[bb++]=i;
else
a[aa++]=i;
}
printf("%d\n",aa);
for(int i=0;i<aa;i++)
printf("%d ",a[i]);
printf("\n%d\n",bb);
for(int i=0;i<bb;i++)
printf("%d ",b[i]);
printf("\n");
return 0;
}