D - NP-Hard Problem

Codeforces Round #360 (Div. 1)

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.

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 k integers — the indices of vertices. Note that because of m ≥ 1, vertex cover cannot be empty.

Example
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：一个无向图，把每一条边的两个结点分到两个集合中，求每个集合中的结点，或无法分

解法：dfs，染色法，只有两种可能，起始结点为1，起始结点相连结点为0，再相邻为1，如果出现两个相邻结点的值相同则输出-1

代码：

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
const int maxn=100010;
int n,m;
vector<int> v[maxn],A,B;
bool vis[maxn];
bool sta[maxn],flag=0;
void dfs(int a,bool s)
{
    if(vis[a]==1&&sta[a]!=s)
        flag=1;
    if(vis[a]==1)
        return;
    if(s)
        A.push_back(a);
    else
        B.push_back(a);
    sta[a]=s;
    vis[a]=1;
    for(int i=0;i<v[a].size();i++){
        dfs(v[a][i],!s);
    }

}
int main()
{
    int x,y;
    cin>>n>>m;
    for(int i=0;i<m;i++){
        cin>>x>>y;
        v[x].push_back(y);
        v[y].push_back(x);
    }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++){
        if(!vis[i])
            dfs(i,0);
    }
    if(flag==1)
        cout<<-1<<endl;
    else{
        cout<<A.size()<<endl;
        for(int i=0;i<A.size();i++){
            if(i!=0)
                cout<<" ";
            cout<<A[i];
        }
        cout<<endl;
        cout<<B.size()<<endl;
        for(int i=0;i<B.size();i++){
            if(i!=0)
                cout<<" ";
            cout<<B[i];
        }
        cout<<endl;
    }
}