Codeforces Round #360 (Div. 2)——C. NP-Hard Problem（BFS染色判二分图）

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.

原来二分图判断是用BFS或DFS染色法，还是对BFS比较熟悉就用BFS了。然而一开始只随便对1这个点进行BFS，并没有考虑到1也许本身就是被舍弃的点，而且数据会出现多个连通分量并存。看了大牛的博客才知道要对每一个节点所在的图都进行判断。难怪一直WA在第15组数据……，也算是学习了二分图的判断方法了

代码：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define MM(x,y) memset(x,y,sizeof(x))
typedef long long LL;
const double PI=acos(-1.0);
const int N=100010;

vector<int>E[N];
int color[N];

void init()
{
	for (int i=0; i<N; i++)
		E[i].clear();
	MM(color,0);
}
bool bfs(int s)
{
	queue<int>Q;
	int i;
	Q.push(s);
	color[s]=1;
	while (!Q.empty())
	{
		int now=Q.front();
		Q.pop();
		int SZ=E[now].size();
		for (i=0; i<SZ; ++i)
		{
			int v=E[now][i];
			if(!color[v])
			{
				color[v]=(color[now]==1?2:1);
				Q.push(v);
			}
			else if(color[v]&&color[v]==color[now])
				return false;			
		}
	}
	return true;
}
int main(void)
{
	int n,m,i,j,k,a,b,c,flag;
	while (~scanf("%d%d",&n,&m))
	{
		init();
		flag=1;
		for (i=0; i<m; i++)
		{
			scanf("%d%d",&a,&b);
			E[a].push_back(b);
			E[b].push_back(a);
		}
		for (i=1; i<=n; i++)
		{
			if(!color[i]&&E[i].size()>0)
			{
				if(!bfs(i))
					flag=0;
			}	
		}
		if(!flag)
			puts("-1");
		else
		{
			int cnta=0,cntb=0;
			vector<int>va,vb;
			for (i=1; i<=n; i++)
			{
				if(color[i]==1)
				{
					va.push_back(i);
					cnta++;
				}
				else if(color[i]==2)
				{
					vb.push_back(i);
					cntb++;
				}
			}
			printf("%d
",cnta);
			for (i=0; i<cnta; i++)
				printf("%d%s",va[i],i==cnta-1?"
":" ");
					
			printf("%d
",cntb);
			for (i=0; i<cntb; i++)
				printf("%d%s",vb[i],i==cntb-1?"
":" ");
		}
	}
	return 0;
}