codeforces C. NP-Hard Problem 二分图的判定

小鱼儿 2022-07-21 01:30 141阅读 0赞

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. ![a815cf6b86613c0097990e1afa181ef245307fa8.png][] or ![5703655b84cc7aaa459b219d9b9b0ed15843c59a.png][] (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 *u**i* and *v**i* (1  ≤  *u**i*,  *v**i*  ≤  *n*), denoting an undirected edge between *u**i* and *v**i*. 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.

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: 把给定的无向图分成两部分, 这两部分不能有重复的点, 每一部分都有每一条边的终点;

分析: 就是要把每一条边连接的两个点划分到两部分里, 就是二分图的判定, 染色法来判断就行;

#include<bits/stdc++.h>
    
    #define inf 0x3f3f3f3f
    #define mem(a,x) memset(a,x,sizeof(a))
    #define F first
    #define S second
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    inline int in()
    {
        int res=0;char c;int f=1;
        while((c=getchar())<'0' || c>'9')if(c=='-')f=-1;
        while(c>='0' && c<='9')res=res*10+c-'0',c=getchar();
        return res*f;
    }
    const int N=100010,MOD=1e9+7;
    
    int color[N],rd[N];
    vector<int> G[N];
    
    bool dfs(int x)
    {
        for(int i=0;i<G[x].size();i++){
            int t=G[x][i];
            if(!color[t]){
                color[t] = 3-color[x];
                if(!dfs(t)) return 0;
            }
            else if(color[t] == color[x]) return 0;
        }
        return 1;
    }
    
    int main()
    {
        int n=in(),m=in();
        for(int i=0;i<m;i++){
            int x=in(),y=in();
            G[x].push_back(y);
            G[y].push_back(x);
            rd[x]++,rd[y]++;
        }
    
        for(int i=1;i<=n;i++){
            if(!color[i] && rd[i]){
                color[i]=1;
                if(!dfs(i)) return 0*puts("-1");
            }
        }
        int k1=0,k2=0;
        for(int i=1;i<=n;i++){
            if(color[i]==1){
                k1++;
            }
            else if(color[i]==2) k2++;
        }
        cout<<k1<<endl;
        for(int i=1;i<=n;i++){
            if(color[i] == 1){
                printf("%d ",i);
            }
        }
        puts("");
        cout<<k2<<endl;
        for(int i=1;i<=n;i++){
            if(color[i] == 2){
                printf("%d ",i);
            }
        }
        return 0;
    }

[a815cf6b86613c0097990e1afa181ef245307fa8.png]: http://codeforces.com/predownloaded/a8/15/a815cf6b86613c0097990e1afa181ef245307fa8.png
[5703655b84cc7aaa459b219d9b9b0ed15843c59a.png]: /images/20220720/9b8fa566881848f38abe7fb85657cf20.png